# Multivariate Normal Distribution¶

## Overview¶

This lecture describes a workhorse in probability theory, statistics, and economics, namely, the multivariate normal distribution.

In this lecture, you will learn formulas for

• the joint distribution of a random vector $x$ of length $N$
• marginal distributions for all subvectors of $x$
• conditional distributions for subvectors of ‘math:x conditional on other subvectors of $x$

We will use the multivariate normal distribution to formulate some classic models:

• a factor analytic model on an intelligence quotient, i.e., IQ
• a factor analytic model or two independent inherent abilities, mathematical and verbal.
• a more general factor analytic model
• PCA as an approximation to a factor analytic model
• time series generated by linear stochastic difference equations

## The Multivariate Normal Distribution¶

This lecture defines a Python class MultivariateNormal to be used to generate marginal and conditional distributions associated with a multivariate normal distribution.

For a multivariate normal distribution it is very convenient that

• conditional expectations equal linear least squares projections
• conditional distributions are characterized by multivariate linear regressions

We apply our Python class to some classic examples.

We will use the following imports:

In :
import numpy as np
from numba import njit
import statsmodels.api as sm
import matplotlib.pyplot as plt
%matplotlib inline


Assume that an $N \times 1$ random vector $z$ has a multivariate normal probability density.

This means that the probability density takes the form

$$f\left(z;\mu,\Sigma\right)=\left(2\pi\right)^{-\left(\frac{N}{2}\right)}\det\left(\Sigma\right)^{-\frac{1}{2}}\exp\left(-.5\left(z-\mu\right)^{\prime}\Sigma^{-1}\left(z-\mu\right)\right)$$

where $\mu=Ez$ is the mean of the random vector $z$ and $\Sigma=E\left(z-\mu\right)\left(z-\mu\right)^\prime$ is the covariance matrix of $z$.

In :
@njit
def f(z, μ, Σ):
"""
The density function of multivariate normal distribution.

Parameters
---------------
z: ndarray(float, dim=2)
random vector, N by 1
μ: ndarray(float, dim=1 or 2)
the mean of z, N by 1
Σ: ndarray(float, dim=2)
the covarianece matrix of z, N by 1
"""

z = np.atleast_2d(z)
μ = np.atleast_2d(μ)
Σ = np.atleast_2d(Σ)

N = z.size

temp1 = np.linalg.det(Σ) ** (-1/2)
temp2 = np.exp(-.5 * (z - μ).T @ np.linalg.inv(Σ) @ (z - μ))

return (2 * np.pi) ** (-N/2) * temp1 * temp2


For some integer $k\in \{2,\dots, N-1\}$, partition $z$ as $z=\left[\begin{array}{c} z_{1}\\ z_{2} \end{array}\right]$, where $z_1$ is an $\left(N-k\right)\times1$ vector and $z_2$ is a $k\times1$ vector.

Let

$$\mu=\left[\begin{array}{c} \mu_{1}\\ \mu_{2} \end{array}\right],\quad\Sigma=\left[\begin{array}{cc} \Sigma_{11} & \Sigma_{12}\\ \Sigma_{21} & \Sigma_{22} \end{array}\right]$$

be corresponding partitions of $\mu$ and $\Sigma$.

The marginal distribution of $z_1$ is

• multivariate normal with mean $\mu_1$ and covariance matrix $\Sigma_{11}$.

The marginal distribution of $z_2$ is

• multivariate normal with mean $\mu_2$ and covariance matrix $\Sigma_{22}$.

The distribution of $z_1$ conditional on $z_2$ is

• multivariate normal with mean
$$\hat{\mu}_1 = \mu_1 + \beta \left(z_2 -\mu_2\right)$$

and covariance matrix

$$\hat{\Sigma}_{11}=\Sigma_{11}-\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}=\Sigma_{11}-\beta\Sigma_{22}\beta^{\prime}$$

where

$$\beta = \Sigma_{12}\Sigma_{22}^{-1}$$

is an $\left(N-k\right) \times k$ matrix of population regression coefficients of $z_1 - \mu_1$ on $z_2 - \mu_2$.

The following class constructs a multivariate normal distribution instance with two methods.

• a method partition computes $\beta$, taking $k$ as an input
• a method cond_dist computes either the distribution of $z_1$ conditional on $z_2$ or the distribution of $z_2$ conditional on $z_1$
In :
class MultivariateNormal:
"""
Class of multivariate normal distribution.

Parameters
----------
μ: ndarray(float, dim=1)
the mean of z, N by 1
Σ: ndarray(float, dim=2)
the covarianece matrix of z, N by 1

Arguments
---------
μ, Σ:
see parameters
μs: list(ndarray(float, dim=1))
list of mean vectors μ1 and μ2 in order
Σs: list(list(ndarray(float, dim=2)))
2 dimensional list of covariance matrices
Σ11, Σ12, Σ21, Σ22 in order
βs: list(ndarray(float, dim=1))
list of regression coefficients β1 and β2 in order
"""

def __init__(self, μ, Σ):
"initialization"
self.μ = np.array(μ)
self.Σ = np.atleast_2d(Σ)

def partition(self, k):
"""
Given k, partition the random vector z into a size k vector z1
and a size N-k vector z2. Partition the mean vector μ into
μ1 and μ2, and the covariance matrix Σ into Σ11, Σ12, Σ21, Σ22
correspondingly. Compute the regression coefficients β1 and β2
using the partitioned arrays.
"""
μ = self.μ
Σ = self.Σ

self.μs = [μ[:k], μ[k:]]
self.Σs = [[Σ[:k, :k], Σ[:k, k:]],
[Σ[k:, :k], Σ[k:, k:]]]

self.βs = [self.Σs @ np.linalg.inv(self.Σs),
self.Σs @ np.linalg.inv(self.Σs)]

def cond_dist(self, ind, z):
"""
Compute the conditional distribution of z1 given z2, or reversely.
Argument ind determines whether we compute the conditional
distribution of z1 (ind=0) or z2 (ind=1).

Returns
---------
μ_hat: ndarray(float, ndim=1)
The conditional mean of z1 or z2.
Σ_hat: ndarray(float, ndim=2)
The conditional covariance matrix of z1 or z2.
"""
β = self.βs[ind]
μs = self.μs
Σs = self.Σs

μ_hat = μs[ind] + β @ (z - μs[1-ind])
Σ_hat = Σs[ind][ind] - β @ Σs[1-ind][1-ind] @ β.T

return μ_hat, Σ_hat


Let’s put this code to work on a suite of examples.

We begin with a simple bivariate example; after that we’ll turn to a trivariate example.

We’ll compute population moments of some conditional distributions using our MultivariateNormal class.

Then for fun we’ll compute sample analogs of the associated population regressions by generating simulations and then computing linear least squares regressions.

We’ll compare those linear least squares regressions for the simulated data to their population counterparts.

## Bivariate Example¶

$$\mu=\left[\begin{array}{c} 0\\ 0 \end{array}\right],\quad\Sigma=\left[\begin{array}{cc} 1 & .2\\ .2 & 1 \end{array}\right]$$
In :
μ = np.array([0., 0.])
Σ = np.array([[1., .2], [.2 ,1.]])

# construction of the multivariate normal instance
multi_normal = MultivariateNormal(μ, Σ)

In :
k = 1 # choose partition

# partition and compute regression coefficients
multi_normal.partition(k)
multi_normal.βs

Out:
array([[0.2]])

Let’s compute the mean and variance of the distribution of $z_1$ conditional on $z_2=5$.

In :
# compute the cond. dist. of z1
ind = 0
z2 = np.array([5.]) # given z2

μ1_hat, Σ1_hat = multi_normal.cond_dist(ind, z2)
print('μ1_hat, Σ1_hat = ', μ1_hat, Σ1_hat)

μ1_hat, Σ1_hat =  [1.] [[0.96]]


Let’s compare the preceding population mean and variance with outcomes from drawing a large sample and then regressing $z_1 - \mu_1$ on $z_2 - \mu_2$.

We know that

$$E z_1 | z_2 = \left(\mu_1 - \beta \mu_2 \right) + \beta z_2$$

which can be arranged to

$$z_1 - \mu_1 = \beta \left( z_2 - \mu_2 \right) + \epsilon,$$

We anticipate that for larger and larger sample sizes, estimated OLS coefficients will converge to $\beta$ and the estimated variance of $\epsilon$ will converge to $\hat{\Sigma}_1$.

In :
n = 1_000_000 # sample size

# simulate multivariate normal random vectors
data = np.random.multivariate_normal(μ, Σ, size=n)
z1_data = data[:, 0]
z2_data = data[:, 1]

# OLS regression
μ1, μ2 = multi_normal.μs
results = sm.OLS(z1_data - μ1, z2_data - μ2).fit()


Let’s compare the preceding population $\beta$ with the OLS sample estimate on $z_2 - \mu_2$

In :
multi_normal.βs, results.params

Out:
(array([[0.2]]), array([0.20022661]))

Let’s compare our population $\hat{\Sigma}_1$ with the degrees-of-freedom adjusted estimate of the variance of $\epsilon$

In :
Σ1_hat, results.resid @ results.resid.T / (n - 1)

Out:
(array([[0.96]]), 0.9609176176231724)

Lastly, let’s compute the estimate of $\hat{E z_1 | z_2}$ and compare it with $\hat{\mu}_1$

In :
μ1_hat, results.predict(z2 - μ2) + μ1

Out:
(array([1.]), array([1.00113307]))

Thus, in each case, for our very large sample size, the sample analogues closely approximate their population counterparts.

These close approximations are foretold by a version of a Law of Large Numbers.

## Trivariate Example¶

Let’s apply our code to a trivariate example.

We’ll specify the mean vector and the covariance matrix as follows.

In :
μ = np.random.random(3)
C = np.random.random((3, 3))
Σ = C @ C.T # positive semi-definite

multi_normal = MultivariateNormal(μ, Σ)

In :
μ, Σ

Out:
(array([0.75398676, 0.28663433, 0.58308708]),
array([[1.48516375, 0.34045372, 0.98710631],
[0.34045372, 0.36640485, 0.44545236],
[0.98710631, 0.44545236, 0.8338027 ]]))
In :
k = 1
multi_normal.partition(k)


Let’s compute the distribution of $z_1$ conditional on $z_{2}=\left[\begin{array}{c} 2\\ 5 \end{array}\right]$.

In :
ind = 0
z2 = np.array([2., 5.])

μ1_hat, Σ1_hat = multi_normal.cond_dist(ind, z2)

In :
n = 1_000_000
data = np.random.multivariate_normal(μ, Σ, size=n)
z1_data = data[:, :k]
z2_data = data[:, k:]

In :
μ1, μ2 = multi_normal.μs
results = sm.OLS(z1_data - μ1, z2_data - μ2).fit()


As above, we compare population and sample regression coefficients, the conditional covariance matrix, and the conditional mean vector in that order.

In :
multi_normal.βs, results.params

Out:
(array([[-1.45531709,  1.96135218]]), array([-1.45558942,  1.96166725]))
In :
Σ1_hat, results.resid @ results.resid.T / (n - 1)

Out:
(array([[0.04456876]]), 0.0446375290720102)
In :
μ1_hat, results.predict(z2 - μ2) + μ1

Out:
(array([6.9236182]), array([6.92454325]))

Once again, sample analogues do a good job of approximating their populations counterparts.

## One Dimensional Intelligence (IQ)¶

Let’s move closer to a real-life example, namely, inferring a one-dimensional measure of intelligence called IQ from a list of test scores.

The $i$th test score $y_i$ equals the sum of an unknown scalar IQ $\theta$ and a random variables $w_{i}$.

$$y_{i} = \theta + \sigma_y w_i, \quad i=1,\dots, n$$

The distribution of IQ’s for a cross-section of people is a normal random variable described by

$$\theta = \mu_{\theta} + \sigma_{\theta} w_{n+1}.$$

We assume the noise in the test scores is IID and not correlated with IQ.

In particular, we assume $\{w_i\}_{i=1}^{n+1}$ are i.i.d. standard normal:

$$\boldsymbol{w}= \left[\begin{array}{c} w_{1}\\ w_{2}\\ \vdots\\ w_{n}\\ w_{n+1} \end{array}\right]\sim N\left(0,I_{n+1}\right)$$

The following system describes the random vector $X$ that interests us:

$$X=\left[\begin{array}{c} y_{1}\\ y_{2}\\ \vdots\\ y_{n}\\ \theta \end{array}\right]=\left[\begin{array}{c} \mu_{\theta}\\ \mu_{\theta}\\ \vdots\\ \mu_{\theta}\\ \mu_{\theta} \end{array}\right]+\left[\begin{array}{ccccc} \sigma_{y} & 0 & \cdots & 0 & \sigma_{\theta}\\ 0 & \sigma_{y} & \cdots & 0 & \sigma_{\theta}\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & \cdots & \sigma_{y} & \sigma_{\theta}\\ 0 & 0 & \cdots & 0 & \sigma_{\theta} \end{array}\right]\left[\begin{array}{c} w_{1}\\ w_{2}\\ \vdots\\ w_{n}\\ w_{n+1} \end{array}\right],$$

or equivalently,

$$X=\mu_{\theta}\boldsymbol{1}_{n+1}+D\boldsymbol{w}$$

where $X = \begin{bmatrix} y \cr \theta \end{bmatrix}$, $\boldsymbol{1}_{n+1}$ is a vector of $1$s of size $n+1$, and $D$ is an $n+1$ by $n+1$ matrix.

Let’s define a Python function that constructs the mean $\mu$ and covariance matrix $\Sigma$ of the random vector $X$ that we know is governed by a multivariate normal distribution.

As arguments, the function takes the number of tests $n$, the mean $\mu_{\theta}$ and the standard deviation $\sigma_\theta$ of the IQ distribution, and the standard deviation of the randomness in test scores $\sigma_{y}$.

In :
def construct_moments_IQ(n, 𝜇𝜃, 𝜎𝜃, 𝜎y):

𝜇_IQ = np.ones(n+1) * 𝜇𝜃

D_IQ = np.zeros((n+1, n+1))
D_IQ[range(n), range(n)] = 𝜎y
D_IQ[:, n] = 𝜎𝜃

Σ_IQ = D_IQ @ D_IQ.T

return 𝜇_IQ, Σ_IQ, D_IQ


Now let’s consider a specific instance of this model.

Assume we have recorded $50$ test scores and we know that $\mu_{\theta}=100$, $\sigma_{\theta}=10$, and $\sigma_{y}=10$.

We can compute the mean vector and covariance matrix of $x$ easily with our construct_moments_IQ function as follows.

In :
n = 50
𝜇𝜃, 𝜎𝜃, 𝜎y = 100., 10., 10.

𝜇_IQ, Σ_IQ, D_IQ = construct_moments_IQ(n, 𝜇𝜃, 𝜎𝜃, 𝜎y)
μ_IQ, Σ_IQ, D_IQ

Out:
(array([100., 100., 100., 100., 100., 100., 100., 100., 100., 100., 100.,
100., 100., 100., 100., 100., 100., 100., 100., 100., 100., 100.,
100., 100., 100., 100., 100., 100., 100., 100., 100., 100., 100.,
100., 100., 100., 100., 100., 100., 100., 100., 100., 100., 100.,
100., 100., 100., 100., 100., 100., 100.]),
array([[200., 100., 100., ..., 100., 100., 100.],
[100., 200., 100., ..., 100., 100., 100.],
[100., 100., 200., ..., 100., 100., 100.],
...,
[100., 100., 100., ..., 200., 100., 100.],
[100., 100., 100., ..., 100., 200., 100.],
[100., 100., 100., ..., 100., 100., 100.]]),
array([[10.,  0.,  0., ...,  0.,  0., 10.],
[ 0., 10.,  0., ...,  0.,  0., 10.],
[ 0.,  0., 10., ...,  0.,  0., 10.],
...,
[ 0.,  0.,  0., ..., 10.,  0., 10.],
[ 0.,  0.,  0., ...,  0., 10., 10.],
[ 0.,  0.,  0., ...,  0.,  0., 10.]]))

We can now use our MultivariateNormal class to construct an instance, then partition the mean vector and covariance matrix as we wish.

We choose k=n so that $z_{1} = y$ and $z_{2} = \theta$.

In :
multi_normal_IQ = MultivariateNormal(μ_IQ, Σ_IQ)

k = n
multi_normal_IQ.partition(k)


Using the generator multivariate_normal, we can make one draw of the random vector from our distribution and then compute the distribution of $\theta$ conditional on our test scores.

Let’s do that and then print out some pertinent quantities.

In :
x = np.random.multivariate_normal(μ_IQ, Σ_IQ)
y = x[:-1] # test scores
𝜃 = x[-1]  # IQ

In :
# the true value
𝜃

Out:
86.06122915425433

The method cond_dist takes test scores as input and returns the conditional normal distribution of the IQ $\theta$.

Note that now $\theta$ is what we denoted as $z_{2}$ in the general case so we need to set ind=1.

In :
ind = 1
multi_normal_IQ.cond_dist(ind, y)

Out:
(array([89.38385826]), array([[1.96078431]]))

The first number is the conditional mean $\hat{\mu}_{\theta}$ and the second is the conditional variance $\hat{\Sigma}_{\theta}$.

How do the additional test scores affect our inferences?

To shed light on this, we compute a sequence of conditional distributions of $\theta$ by varying the number of test scores in the conditioning set from $1$ to $n$.

We’ll make a pretty graph showing how our judgment of the person’s IQ change as more test results come in.

In :
# array for containing moments
μ𝜃_hat_arr = np.empty(n)
Σ𝜃_hat_arr = np.empty(n)

# loop over number of test scores
for i in range(1, n+1):
# construction of multivariate normal distribution instance
𝜇_IQ_i, Σ_IQ_i, D_IQ_i = construct_moments_IQ(i, 𝜇𝜃, 𝜎𝜃, 𝜎y)
multi_normal_IQ_i = MultivariateNormal(μ_IQ_i, Σ_IQ_i)

# partition and compute conditional distribution
multi_normal_IQ_i.partition(i)
scores_i = y[:i]
μ𝜃_hat_i, Σ𝜃_hat_i = multi_normal_IQ_i.cond_dist(1, scores_i)

# store the results
μ𝜃_hat_arr[i-1] = μ𝜃_hat_i
Σ𝜃_hat_arr[i-1] = Σ𝜃_hat_i[0, 0]

# transform variance to standard deviation
𝜎𝜃_hat_arr = np.sqrt(Σ𝜃_hat_arr)

In :
μ𝜃_hat_lower = μ𝜃_hat_arr - 1.96 * 𝜎𝜃_hat_arr
μ𝜃_hat_higher = μ𝜃_hat_arr + 1.96 * 𝜎𝜃_hat_arr

plt.hlines(𝜃, 1, n+1, ls='--', label='true $𝜃$')
plt.plot(range(1, n+1), μ𝜃_hat_arr, color='b', label='$\hat{μ}_{𝜃}$')
plt.plot(range(1, n+1), μ𝜃_hat_lower, color='b', ls='--')
plt.plot(range(1, n+1), μ𝜃_hat_higher, color='b', ls='--')
plt.fill_between(range(1, n+1), μ𝜃_hat_lower, μ𝜃_hat_higher,
color='b', alpha=0.2, label='95%')

plt.xlabel('number of test scores')
plt.ylabel('$\hat{𝜃}$')
plt.legend()

plt.show() The solid blue line in the plot above shows $\hat{\mu}_{\theta}$ as function of the number of test scores that we have recorded and conditioned on.

The blue area shows the span that comes from adding or deducing $1.96 \hat{\sigma}_{\theta}$ from $\hat{\mu}_{\theta}$.

Therefore, $95\%$ of the probability mass of the conditional distribution falls in this range.

The value of the random $\theta$ that we drew is shown by the black dotted line.

As more and more test scores come in, our estimate of the person’s $\theta$ become more and more reliable.

By staring at the changes in the conditional distributions, we see that adding more test scores makes $\hat{\theta}$ settle down and approach $\theta$.

Thus, each $y_{i}$ adds information about $\theta$.

If we drove the number of tests $n \rightarrow + \infty$, the conditional standard deviation $\hat{\sigma}_{\theta}$ would converge to $0$ at the rate $\frac{1}{n^{.5}}$.

## Another representation¶

By using a different representation, let’s look at things from a different perspective.

We can represent the random vector $X$ defined above as

$$X = \mu_{\theta} \boldsymbol{1}_{n+1} + C \epsilon, \quad \epsilon \sim N\left(0, I\right)$$

where $C$ is a lower triangular Cholesky factor of $\Sigma$ so that

$$\Sigma \equiv DD^{\prime} = C C^\prime$$

and

$$E \epsilon \epsilon' = I .$$

It follows that

$$\epsilon \sim N(0, I) .$$

Let $G=C^{-1}$; $G$ is also lower triangular.

We can compute $\epsilon$ from the formula

$$\epsilon = G \left( X - \mu_{\theta} \boldsymbol{1}_{n+1} \right)$$

This formula confirms that the orthonormal vector $\epsilon$ contains the same information as the non-orthogonal vector $\left( X - \mu_{\theta} \boldsymbol{1}_{n+1} \right)$.

We can say that $\epsilon$ is an orthogonal basis for $\left( X - \mu_{\theta} \boldsymbol{1}_{n+1} \right)$.

Let $c_{i}$ be the $i$th element in the last row of $C$.

Then we can write

$$\theta = \mu_{\theta} + c_1 \epsilon_1 + c_2 \epsilon_2 + \dots + c_n \epsilon_n + c_{n+1} \epsilon_{n+1} \tag{1}$$

The mutual orthogonality of the $\epsilon_i$’s provides us an informative way to interpret them in light of equation (1).

Thus, relative to what is known from tests $i=1, \ldots, n-1$, $c_i \epsilon_i$ is the amount of new information about $\theta$ brought by the test number $i$.

Here new information means surprise or what could not be predicted from earlier information.

Formula (1) also provides us with an enlightening way to express conditional means and conditional variances that we computed earlier.

In particular,

$$E\left[\theta \mid y_1, \dots, y_k\right] = \mu_{\theta} + c_1 \epsilon_1 + \dots + c_k \epsilon_k$$

and

$$Var\left(\theta \mid y_1, \dots, y_k\right) = c^2_{k+1} + c^2_{k+2} + \dots + c^2_{n+1}.$$
In :
C = np.linalg.cholesky(Σ_IQ)
G = np.linalg.inv(C)

𝜖 = G @ (x - 𝜇𝜃)

In :
c𝜖 = C[n, :] * 𝜖

# compute the sequence of μ𝜃 and Σ𝜃 conditional on y1, y2, ..., yk
μ𝜃_hat_arr_C = np.array([np.sum(c𝜖[:k+1]) for k in range(n)]) + 𝜇𝜃
Σ𝜃_hat_arr_C = np.array([np.sum(C[n, i+1:n+1] ** 2) for i in range(n)])


To confirm that these formulas give the same answers that we computed earlier, we can compare the means and variances of $\theta$ conditional on $\{y_i\}_{i=1}^k$ with what we obtained above using the formulas implemented in the class MultivariateNormal built on our original representation of conditional distributions for multivariate normal distributions.

In :
# conditional mean
np.max(np.abs(μ𝜃_hat_arr - μ𝜃_hat_arr_C)) < 1e-10

Out:
True
In :
# conditional variance
np.max(np.abs(Σ𝜃_hat_arr - Σ𝜃_hat_arr_C)) < 1e-10

Out:
True

## Magic of the Cholesky factorization¶

Evidently, the Cholesky factorization is automatically computing the population regression coefficients and associated statistics that are produced by our MultivariateNormal class.

And they are doing it recursively.

Indeed, in formula (1),

• the random variable $c_i \epsilon_i$ is information about $\theta$ that is not contained by the information in $\epsilon_1, \epsilon_2, \ldots, \epsilon_{i-1}$
• the coefficient $c_i$ is the simple population regression coefficient of $\theta - \mu_\theta$ on $\epsilon_i$

## Math and Verbal Components of Intelligence¶

We can alter the preceding example to be more realistic.

There is ample evidence that IQ is not a scalar.

Some people are good in math skills but poor in language skills.

Other people are good in language skills but poor in math skills.

So now we shall assume that there are two dimensions of IQ, $\theta$ and $\eta$.

These determine average performances in math and language tests, respectively.

We observe math scores $\{y_i\}_{i=1}^{n}$ and language scores $\{y_i\}_{i=n+1}^{2n}$.

When $n=2$, we assume that outcomes are draws from a multivariate normal distribution with representation

$$X=\left[\begin{array}{c} y_{1}\\ y_{2}\\ y_{3}\\ y_{4}\\ \theta\\ \eta \end{array}\right]=\left[\begin{array}{c} \mu_{\theta}\\ \mu_{\theta}\\ \mu_{\eta}\\ \mu_{\eta}\\ \mu_{\theta}\\ \mu_{\eta} \end{array}\right]+\left[\begin{array}{cccccc} \sigma_{y} & 0 & 0 & 0 & \sigma_{\theta} & 0\\ 0 & \sigma_{y} & 0 & 0 & \sigma_{\theta} & 0\\ 0 & 0 & \sigma_{y} & 0 & 0 & \sigma_{\eta}\\ 0 & 0 & 0 & \sigma_{y} & 0 & \sigma_{\eta}\\ 0 & 0 & 0 & 0 & \sigma_{\theta} & 0\\ 0 & 0 & 0 & 0 & 0 & \sigma_{\eta} \end{array}\right]\left[\begin{array}{c} w_{1}\\ w_{2}\\ w_{3}\\ w_{4}\\ w_{5}\\ w_{6} \end{array}\right]$$

where $w \begin{bmatrix} w_1 \cr w_2 \cr \vdots \cr w_6 \end{bmatrix}$ is a standard normal random vector.

We construct a Python function construct_moments_IQ2d to construct the mean vector and covariance matrix of the joint normal distribution.

In :
def construct_moments_IQ2d(n, 𝜇𝜃, 𝜎𝜃, 𝜇𝜂, 𝜎𝜂, 𝜎y):

𝜇_IQ2d = np.empty(2*(n+1))
𝜇_IQ2d[:n] = 𝜇𝜃
𝜇_IQ2d[2*n] = 𝜇𝜃
𝜇_IQ2d[n:2*n] = 𝜇𝜂
𝜇_IQ2d[2*n+1] = 𝜇𝜂

D_IQ2d = np.zeros((2*(n+1), 2*(n+1)))
D_IQ2d[range(2*n), range(2*n)] = 𝜎y
D_IQ2d[:n, 2*n] = 𝜎𝜃
D_IQ2d[2*n, 2*n] = 𝜎𝜃
D_IQ2d[n:2*n, 2*n+1] = 𝜎𝜂
D_IQ2d[2*n+1, 2*n+1] = 𝜎𝜂

Σ_IQ2d = D_IQ2d @ D_IQ2d.T

return 𝜇_IQ2d, Σ_IQ2d, D_IQ2d


Let’s put the function to work.

In :
n = 2
# mean and variance of 𝜃, 𝜂, and y
𝜇𝜃, 𝜎𝜃, 𝜇𝜂, 𝜎𝜂, 𝜎y = 100., 10., 100., 10, 10

𝜇_IQ2d, Σ_IQ2d, D_IQ2d = construct_moments_IQ2d(n, 𝜇𝜃, 𝜎𝜃, 𝜇𝜂, 𝜎𝜂, 𝜎y)
𝜇_IQ2d, Σ_IQ2d, D_IQ2d

Out:
(array([100., 100., 100., 100., 100., 100.]),
array([[200., 100.,   0.,   0., 100.,   0.],
[100., 200.,   0.,   0., 100.,   0.],
[  0.,   0., 200., 100.,   0., 100.],
[  0.,   0., 100., 200.,   0., 100.],
[100., 100.,   0.,   0., 100.,   0.],
[  0.,   0., 100., 100.,   0., 100.]]),
array([[10.,  0.,  0.,  0., 10.,  0.],
[ 0., 10.,  0.,  0., 10.,  0.],
[ 0.,  0., 10.,  0.,  0., 10.],
[ 0.,  0.,  0., 10.,  0., 10.],
[ 0.,  0.,  0.,  0., 10.,  0.],
[ 0.,  0.,  0.,  0.,  0., 10.]]))
In :
# take one draw
x = np.random.multivariate_normal(μ_IQ2d, Σ_IQ2d)
y1 = x[:n]
y2 = x[n:2*n]
𝜃 = x[2*n]
𝜂 = x[2*n+1]

# the true values
𝜃, 𝜂

Out:
(90.76860453625066, 119.59870646081532)

We first compute the joint normal distribution of $\left(\theta, \eta\right)$.

In :
multi_normal_IQ2d = MultivariateNormal(𝜇_IQ2d, Σ_IQ2d)

k = 2*n # the length of data vector
multi_normal_IQ2d.partition(k)

multi_normal_IQ2d.cond_dist(1, [*y1, *y2])

Out:
(array([ 96.00517808, 116.92009856]), array([[33.33333333,  0.        ],
[ 0.        , 33.33333333]]))

Now let’s compute distributions of $\theta$ and $\mu$ separately conditional on various subsets of test scores.

It will be fun to compare outcomes with the help of an auxiliary function cond_dist_IQ2d that we now construct.

In :
def cond_dist_IQ2d(𝜇, Σ, data):

n = len(𝜇)

multi_normal = MultivariateNormal(𝜇, Σ)
multi_normal.partition(n-1)
𝜇_hat, Σ_hat = multi_normal.cond_dist(1, data)

return 𝜇_hat, Σ_hat


Let’s see how things work for an example.

In :
for indices, IQ, conditions in [([*range(2*n), 2*n], '𝜃', 'y1, y2, y3, y4'),
([*range(n), 2*n], '𝜃', 'y1, y2'),
([*range(n, 2*n), 2*n], '𝜃', 'y3, y4'),
([*range(2*n), 2*n+1], '𝜂', 'y1, y2, y3, y4'),
([*range(n), 2*n+1], '𝜂', 'y1, y2'),
([*range(n, 2*n), 2*n+1], '𝜂', 'y3, y4')]:

𝜇_hat, Σ_hat = cond_dist_IQ2d(𝜇_IQ2d[indices], Σ_IQ2d[indices][:, indices], x[indices[:-1]])
print(f'The mean and variance of {IQ} conditional on {conditions: <15} are ' +
f'{𝜇_hat:1.2f} and {Σ_hat[0, 0]:1.2f} respectively')

The mean and variance of 𝜃 conditional on y1, y2, y3, y4  are 96.01 and 33.33 respectively
The mean and variance of 𝜃 conditional on y1, y2          are 96.01 and 33.33 respectively
The mean and variance of 𝜃 conditional on y3, y4          are 100.00 and 100.00 respectively
The mean and variance of 𝜂 conditional on y1, y2, y3, y4  are 116.92 and 33.33 respectively
The mean and variance of 𝜂 conditional on y1, y2          are 100.00 and 100.00 respectively
The mean and variance of 𝜂 conditional on y3, y4          are 116.92 and 33.33 respectively


Evidently, math tests provide no information about $\mu$ and language tests provide no information about $\eta$.

## Univariate Time Series Analysis¶

We can use the multivariate normal distribution and a little matrix algebra to present foundations of univariate linear time series analysis.

Let $x_t, y_t, v_t, w_{t+1}$ each be scalars for $t \geq 0$.

Consider the following model:

\begin{aligned} x_0 & \sim N\left(0, \sigma_0^2\right) \\ x_{t+1} & = a x_{t} + b w_{t+1}, \quad w_{t+1} \sim N\left(0, 1\right), t \geq 0 \\ y_{t} & = c x_{t} + d v_{t}, \quad v_{t} \sim N\left(0, 1\right), t \geq 0 \end{aligned}

We can compute the moments of $x_{t}$

1. $E x_{t+1}^2 = a^2 E x_{t}^2 + b^2, t \geq 0$, where $E x_{0}^2 = \sigma_{0}^2$
2. $E x_{t+j} x_{t} = a^{j} E x_{t}^2, \forall t \ \forall j$

Given some $T$, we can formulate the sequence $\{x_{t}\}_{t=0}^T$ as a random vector

$$X=\left[\begin{array}{c} x_{0}\\ x_{1}\\ \vdots\\ x_{T} \end{array}\right]$$

and the covariance matrix $\Sigma_{x}$ can be constructed using the moments we have computed above.

Similarly, we can define

$$Y=\left[\begin{array}{c} y_{0}\\ y_{1}\\ \vdots\\ y_{T} \end{array}\right], \quad v=\left[\begin{array}{c} v_{0}\\ v_{1}\\ \vdots\\ v_{T} \end{array}\right]$$

and therefore

$$Y = C X + D V$$

where $C$ and $D$ are both diagonal matrices with constant $c$ and $d$ as diagonal respectively.

Consequently, the covariance matrix of $Y$ is

$$\Sigma_{y} = E Y Y^{\prime} = C \Sigma_{x} C^{\prime} + D D^{\prime}$$

By stacking $X$ and $Y$, we can write

$$Z=\left[\begin{array}{c} X\\ Y \end{array}\right]$$

and

$$\Sigma_{z} = EZZ^{\prime}=\left[\begin{array}{cc} \Sigma_{x} & \Sigma_{x}C^{\prime}\\ C\Sigma_{x} & \Sigma_{y} \end{array}\right]$$

Thus, the stacked sequences $\{x_{t}\}_{t=0}^T$ and $\{y_{t}\}_{t=0}^T$ jointly follow the multivariate normal distribution $N\left(0, \Sigma_{z}\right)$.

In :
# as an example, consider the case where T = 3
T = 3

In :
# variance of the initial distribution x_0
𝜎0 = 1.

# parameters of the equation system
a = .9
b = 1.
c = 1.0
d = .05

In :
# construct the covariance matrix of X
Σx = np.empty((T+1, T+1))

Σx[0, 0] = 𝜎0 ** 2
for i in range(T):
Σx[i, i+1:] = Σx[i, i] * a ** np.arange(1, T+1-i)
Σx[i+1:, i] = Σx[i, i+1:]

Σx[i+1, i+1] = a ** 2 * Σx[i, i] + b ** 2

In :
Σx

Out:
array([[1.      , 0.9     , 0.81    , 0.729   ],
[0.9     , 1.81    , 1.629   , 1.4661  ],
[0.81    , 1.629   , 2.4661  , 2.21949 ],
[0.729   , 1.4661  , 2.21949 , 2.997541]])
In :
# construct the covariance matrix of Y
C = np.eye(T+1) * c
D = np.eye(T+1) * d

Σy = C @ Σx @ C.T + D @ D.T

In :
# construct the covariance matrix of Z
Σz = np.empty((2*(T+1), 2*(T+1)))

Σz[:T+1, :T+1] = Σx
Σz[:T+1, T+1:] = Σx @ C.T
Σz[T+1:, :T+1] = C @ Σx
Σz[T+1:, T+1:] = Σy

In :
Σz

Out:
array([[1.      , 0.9     , 0.81    , 0.729   , 1.      , 0.9     ,
0.81    , 0.729   ],
[0.9     , 1.81    , 1.629   , 1.4661  , 0.9     , 1.81    ,
1.629   , 1.4661  ],
[0.81    , 1.629   , 2.4661  , 2.21949 , 0.81    , 1.629   ,
2.4661  , 2.21949 ],
[0.729   , 1.4661  , 2.21949 , 2.997541, 0.729   , 1.4661  ,
2.21949 , 2.997541],
[1.      , 0.9     , 0.81    , 0.729   , 1.0025  , 0.9     ,
0.81    , 0.729   ],
[0.9     , 1.81    , 1.629   , 1.4661  , 0.9     , 1.8125  ,
1.629   , 1.4661  ],
[0.81    , 1.629   , 2.4661  , 2.21949 , 0.81    , 1.629   ,
2.4686  , 2.21949 ],
[0.729   , 1.4661  , 2.21949 , 2.997541, 0.729   , 1.4661  ,
2.21949 , 3.000041]])
In :
# construct the mean vector of Z
𝜇z = np.zeros(2*(T+1))


The following Python code lets us sample random vectors $X$ and $Y$.

This is going to be very useful for doing the conditioning to be used in the fun exercises below.

In :
z = np.random.multivariate_normal(𝜇z, Σz)

x = z[:T+1]
y = z[T+1:]


### Smoothing Example¶

This is an instance of a classic smoothing calculation whose purpose is to compute $E X \mid Y$.

An interpretation of this example is

• $X$ is a random sequence of hidden Markov state variables $x_t$
• $Y$ is a sequence of observed signals $y_t$ bearing information about the hidden state
In :
# construct a MultivariateNormal instance
multi_normal_ex1 = MultivariateNormal(𝜇z, Σz)
x = z[:T+1]
y = z[T+1:]

In :
# partition Z into X and Y
multi_normal_ex1.partition(T+1)

In :
# compute the conditional mean and covariance matrix of X given Y=y

print("X = ", x)
print("Y = ", y)
print(" E [ X | Y] = ", )

multi_normal_ex1.cond_dist(0, y)

X =  [-0.73092347 -0.80558984 -1.23375625 -2.1096767 ]
Y =  [-0.69809204 -0.81897485 -1.20389596 -2.16274472]
E [ X | Y] =

Out:
(array([-0.69678307, -0.81954574, -1.20514678, -2.16005616]),
array([[2.48875094e-03, 5.57449314e-06, 1.24861729e-08, 2.80235835e-11],
[5.57449314e-06, 2.48876343e-03, 5.57452116e-06, 1.25113941e-08],
[1.24861729e-08, 5.57452116e-06, 2.48876346e-03, 5.58575339e-06],
[2.80235835e-11, 1.25113941e-08, 5.58575339e-06, 2.49377812e-03]]))

### Filtering Exercise¶

Compute $E\left[x_{t} \mid y_{t-1}, y_{t-2}, \dots, y_{0}\right]$.

To do so, we need to first construct the mean vector and the covariance matrix of the subvector $\left[x_{t}, y_{0}, \dots, y_{t-2}, y_{t-1}\right]$.

For example, let’s say that we want the conditional distribution of $x_{3}$.

In :
t = 3

In :
# mean of the subvector
sub_𝜇z = np.zeros(t+1)

# covariance matrix of the subvector
sub_Σz = np.empty((t+1, t+1))

sub_Σz[0, 0] = Σz[t, t] # x_t
sub_Σz[0, 1:] = Σz[t, T+1:T+t+1]
sub_Σz[1:, 0] = Σz[T+1:T+t+1, t]
sub_Σz[1:, 1:] = Σz[T+1:T+t+1, T+1:T+t+1]

In :
sub_Σz

Out:
array([[2.997541, 0.729   , 1.4661  , 2.21949 ],
[0.729   , 1.0025  , 0.9     , 0.81    ],
[1.4661  , 0.9     , 1.8125  , 1.629   ],
[2.21949 , 0.81    , 1.629   , 2.4686  ]])
In :
multi_normal_ex2 = MultivariateNormal(sub_𝜇z, sub_Σz)
multi_normal_ex2.partition(1)

In :
sub_y = y[:t]

multi_normal_ex2.cond_dist(0, sub_y)

Out:
(array([-1.08245979]), array([[1.00201996]]))

### Prediction Exercise¶

Compute $E\left[y_{t} \mid y_{t-j}, \dots, y_{0} \right]$.

As what we did in exercise 2, we will construct the mean vector and covariance matrix of the subvector $\left[y_{t}, y_{0}, \dots, y_{t-j-1}, y_{t-j} \right]$.

For example, we take a case in which $t=3$ and $j=2$.

In :
t = 3
j = 2

In :
sub_𝜇z = np.zeros(t-j+2)
sub_Σz = np.empty((t-j+2, t-j+2))

sub_Σz[0, 0] = Σz[T+t+1, T+t+1]
sub_Σz[0, 1:] = Σz[T+t+1, T+1:T+t-j+2]
sub_Σz[1:, 0] = Σz[T+1:T+t-j+2, T+t+1]
sub_Σz[1:, 1:] = Σz[T+1:T+t-j+2, T+1:T+t-j+2]

In :
sub_Σz

Out:
array([[3.000041, 0.729   , 1.4661  ],
[0.729   , 1.0025  , 0.9     ],
[1.4661  , 0.9     , 1.8125  ]])
In :
multi_normal_ex3 = MultivariateNormal(sub_𝜇z, sub_Σz)
multi_normal_ex3.partition(1)

In :
sub_y = y[:t-j+1]

multi_normal_ex3.cond_dist(0, sub_y)

Out:
(array([-0.66298205]), array([[1.81413617]]))

### Constructing a Wold Representation¶

Now we’ll apply Cholesky decomposition to decompose $\Sigma_{y}=H H^{\prime}$ and form

$$\epsilon = H^{-1} Y.$$

Then we can represent $y_{t}$ as

$$y_{t} = h_{t,t} \epsilon_{t} + h_{t,t-1} \epsilon_{t-1} + \dots + h_{t,0} \epsilon_{0}.$$
In :
H = np.linalg.cholesky(Σy)

H

Out:
array([[1.00124922, 0.        , 0.        , 0.        ],
[0.8988771 , 1.00225743, 0.        , 0.        ],
[0.80898939, 0.89978675, 1.00225743, 0.        ],
[0.72809046, 0.80980808, 0.89978676, 1.00225743]])
In :
𝜖 = np.linalg.inv(H) @ y

𝜖

Out:
array([-0.69722105, -0.19182577, -0.46619683, -1.07785175])
In :
y

Out:
array([-0.69809204, -0.81897485, -1.20389596, -2.16274472])

This example is an instance of what is known as a Wold representation in time series analysis.

## Classic Factor Analysis Model¶

The factor analysis model widely used in psychology and other fields can be represented as

$$Y = \Lambda f + U$$

where

1. $Y$ is $n \times 1$ random vector, $E U U^{\prime} = D$ is a diagonal matrix,
2. $\Lambda$ is $n \times k$ coefficient matrix,
3. $f$ is $k \times 1$ random vector, $E f f^{\prime} = I$,
4. $U$ is $n \times 1$ random vector, and $U \perp f$.
5. It is presumed that $k$ is small relative to $n$; often $k$ is only $1$ or $2$, as in our IQ examples.

This implies that

$$\Sigma_y = E Y Y^{\prime} = \Lambda \Lambda^{\prime} + D \\ E Y f^{\prime} = \Lambda \\ E f Y^{\prime} = \Lambda^{\prime}$$

Thus, the covariance matrix $\Sigma_Y$ is the sum of a diagonal matrix $D$ and a positive semi-definite matrix $\Lambda \Lambda^{\prime}$ of rank $k$.

This means that all covariances among the $n$ components of the $Y$ vector are intermediated by their common dependencies on the $k<$ factors.

Form

$$Z=\left(\begin{array}{c} f\\ Y \end{array}\right)$$

the covariance matrix of the expanded random vector $Z$ can be computed as

$$\Sigma_{z} = EZZ^{\prime}=\left(\begin{array}{cc} I & \Lambda^{\prime}\\ \Lambda & \Lambda\Lambda^{\prime}+D \end{array}\right)$$

In the following, we first construct the mean vector and the covariance matrix for the case where $N=10$ and $k=2$.

In :
N = 10
k = 2


We set the coefficient matrix $\Lambda$ and the covariance matrix of $U$ to be

$$\Lambda=\left(\begin{array}{cc} 1 & 0\\ \vdots & \vdots\\ 1 & 0\\ 0 & 1\\ \vdots & \vdots\\ 0 & 1 \end{array}\right),\quad D=\left(\begin{array}{cccc} \sigma_{u}^{2} & 0 & \cdots & 0\\ 0 & \sigma_{u}^{2} & \cdots & 0\\ \vdots & \vdots & \vdots & \vdots\\ 0 & 0 & \cdots & \sigma_{u}^{2} \end{array}\right)$$

where the first half of the first column of $\Lambda$ is filled with $1$s and $0$s for the rest half, and symmetrically for the second column. $D$ is a diagonal matrix with parameter $\sigma_{u}^{2}$ on the diagonal.

In :
Λ = np.zeros((N, k))
Λ[:N//2, 0] = 1
Λ[N//2:, 1] = 1

𝜎u = .5
D = np.eye(N) * 𝜎u ** 2

In :
# compute Σy
Σy = Λ @ Λ.T + D


We can now construct the mean vector and the covariance matrix for $Z$.

In :
𝜇z = np.zeros(k+N)

Σz = np.empty((k+N, k+N))

Σz[:k, :k] = np.eye(k)
Σz[:k, k:] = Λ.T
Σz[k:, :k] = Λ
Σz[k:, k:] = Σy

In :
z = np.random.multivariate_normal(𝜇z, Σz)

f = z[:k]
y = z[k:]

In :
multi_normal_factor = MultivariateNormal(𝜇z, Σz)
multi_normal_factor.partition(k)


Let’s compute the conditional distribution of the hidden factor $f$ on the observations $Y$, namely, $f \mid Y=y$.

In :
multi_normal_factor.cond_dist(0, y)

Out:
(array([-0.20573797,  0.01092524]), array([[0.04761905, 0.        ],
[0.        , 0.04761905]]))

We can verify that the conditional mean $E \left[f \mid Y=y\right] = B Y$ where $B = \Lambda^{\prime} \Sigma_{y}^{-1}$.

In :
B = Λ.T @ np.linalg.inv(Σy)

B @ y

Out:
array([-0.20573797,  0.01092524])

Similarly, we can compute the conditional distribution $Y \mid f$.

In :
multi_normal_factor.cond_dist(1, f)

Out:
(array([-0.15011744, -0.15011744, -0.15011744, -0.15011744, -0.15011744,
0.04109813,  0.04109813,  0.04109813,  0.04109813,  0.04109813]),
array([[0.25, 0.  , 0.  , 0.  , 0.  , 0.  , 0.  , 0.  , 0.  , 0.  ],
[0.  , 0.25, 0.  , 0.  , 0.  , 0.  , 0.  , 0.  , 0.  , 0.  ],
[0.  , 0.  , 0.25, 0.  , 0.  , 0.  , 0.  , 0.  , 0.  , 0.  ],
[0.  , 0.  , 0.  , 0.25, 0.  , 0.  , 0.  , 0.  , 0.  , 0.  ],
[0.  , 0.  , 0.  , 0.  , 0.25, 0.  , 0.  , 0.  , 0.  , 0.  ],
[0.  , 0.  , 0.  , 0.  , 0.  , 0.25, 0.  , 0.  , 0.  , 0.  ],
[0.  , 0.  , 0.  , 0.  , 0.  , 0.  , 0.25, 0.  , 0.  , 0.  ],
[0.  , 0.  , 0.  , 0.  , 0.  , 0.  , 0.  , 0.25, 0.  , 0.  ],
[0.  , 0.  , 0.  , 0.  , 0.  , 0.  , 0.  , 0.  , 0.25, 0.  ],
[0.  , 0.  , 0.  , 0.  , 0.  , 0.  , 0.  , 0.  , 0.  , 0.25]]))

It can be verified that the mean is $\Lambda I^{-1} f = \Lambda f$.

In :
Λ @ f

Out:
array([-0.15011744, -0.15011744, -0.15011744, -0.15011744, -0.15011744,
0.04109813,  0.04109813,  0.04109813,  0.04109813,  0.04109813])

## PCA as Approximation to Factor Analytic Model¶

For fun, let’s apply a Principal Components Analysis (PCA) decomposition to a covariance matrix $\Sigma_y$ that in fact is governed by our factor-analytic model.

Technically, this means that the PCA model is misspecified. (Can you explain why?)

Nevertheless, this exercise will let us study how well the first two principal components from a PCA can approximate the conditional expectations $E f_i | Y$ for our two factors $f_i$, $i=1,2$ for the factor analytic model that we have assumed truly governs the data on $Y$ we have generated.

So we compute the PCA decomposition

$$\Sigma_{y} = P \tilde{\Lambda} P^{\prime}$$

where $\tilde{\Lambda}$ is a diagonal matrix.

We have

$$Y = P \epsilon$$

and

$$\epsilon = P^\prime Y$$

Note that we will arrange the eigenvectors in $P$ in the descending order of eigenvalues.

In :
𝜆_tilde, P = np.linalg.eigh(Σy)

# arrange the eigenvectors by eigenvalues
ind = sorted(range(N), key=lambda x: 𝜆_tilde[x], reverse=True)

P = P[:, ind]
𝜆_tilde = 𝜆_tilde[ind]
Λ_tilde = np.diag(𝜆_tilde)

print('𝜆_tilde =', 𝜆_tilde)

𝜆_tilde = [5.25 5.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25]

In :
# verify the orthogonality of eigenvectors
np.abs(P @ P.T - np.eye(N)).max()

Out:
2.7639210781816965e-16
In :
# verify the eigenvalue decomposition is correct
P @ Λ_tilde @ P.T

Out:
array([[1.25, 1.  , 1.  , 1.  , 1.  , 0.  , 0.  , 0.  , 0.  , 0.  ],
[1.  , 1.25, 1.  , 1.  , 1.  , 0.  , 0.  , 0.  , 0.  , 0.  ],
[1.  , 1.  , 1.25, 1.  , 1.  , 0.  , 0.  , 0.  , 0.  , 0.  ],
[1.  , 1.  , 1.  , 1.25, 1.  , 0.  , 0.  , 0.  , 0.  , 0.  ],
[1.  , 1.  , 1.  , 1.  , 1.25, 0.  , 0.  , 0.  , 0.  , 0.  ],
[0.  , 0.  , 0.  , 0.  , 0.  , 1.25, 1.  , 1.  , 1.  , 1.  ],
[0.  , 0.  , 0.  , 0.  , 0.  , 1.  , 1.25, 1.  , 1.  , 1.  ],
[0.  , 0.  , 0.  , 0.  , 0.  , 1.  , 1.  , 1.25, 1.  , 1.  ],
[0.  , 0.  , 0.  , 0.  , 0.  , 1.  , 1.  , 1.  , 1.25, 1.  ],
[0.  , 0.  , 0.  , 0.  , 0.  , 1.  , 1.  , 1.  , 1.  , 1.25]])
In :
𝜖 = P.T @ y

print("𝜖 = ", 𝜖)

𝜖 =  [-0.4830463   0.02565105 -0.37730731  0.13031767 -0.18464087 -0.75847335
-0.21654499 -0.11248043 -0.31593783  0.26896121]

In :
# print the values of the two factors

print('f = ', f)

f =  [-0.15011744  0.04109813]


Below we’ll plot several things

• the $N$ values of $y$
• the $N$ values of the principal components $\epsilon$
• the value of the first factor $f_1$ plotted only for the first $N/2$ observations of $y$ for which it receives a non-zero loading in $\Lambda$
• the value of the second factor $f_2$ plotted only for the final $N/2$ observations for which it receives a non-zero loading in $\Lambda$
In :
plt.scatter(range(N), y, label='y')
plt.scatter(range(N), 𝜖, label='$\epsilon$')
plt.hlines(f, 0, N//2-1, ls='--', label='$f_{1}$')
plt.hlines(f, N//2, N-1, ls='-.', label='$f_{2}$')
plt.legend()

plt.show() Consequently, the first two $\epsilon_{j}$ correspond to the largest two eigenvalues.

Let’s look at them, after which we’ll look at $E f | y = B y$

In :
𝜖[:2]

Out:
array([-0.4830463 ,  0.02565105])
In :
# compare with Ef|y
B @ y

Out:
array([-0.20573797,  0.01092524])

The fraction of variance in $y_{t}$ explained by the first two principal component can be computed as below.

In :
𝜆_tilde[:2].sum() / 𝜆_tilde.sum()

Out:
0.8400000000000001

Compute

$$\hat{Y} = P_{j} \epsilon_{j} + P_{k} \epsilon_{k}$$

where $P_{j}$ and $P_{k}$ correspond to the largest two eigenvalues.

In :
y_hat = P[:, :2] @ 𝜖[:2]


In this example, it turns out that the projection $\hat{Y}$ of $Y$ on the first two principal components does a good job of approximating $Ef \mid y$.

We confirm this in the following plot of $f$, $E y \mid f$, $E f \mid y$, and $\hat{y}$ on the coordinate axis versus $y$ on the ordinate axis.

In :
plt.scatter(range(N), Λ @ f, label='$Ey|f$')
plt.scatter(range(N), y_hat, label='$\hat{y}$')
plt.hlines(f, 0, N//2-1, ls='--', label='$f_{1}$')
plt.hlines(f, N//2, N-1, ls='-.', label='$f_{2}$')

Efy = B @ y
plt.hlines(Efy, 0, N//2-1, ls='--', color='b', label='$Ef_{1}|y$')
plt.hlines(Efy, N//2, N-1, ls='-.', color='b', label='$Ef_{2}|y$')
plt.legend()

plt.show() The covariance matrix of $\hat{Y}$ can be computed by first constructing the covariance matrix of $\epsilon$ and then use the upper left block for $\epsilon_{1}$ and $\epsilon_{2}$.

In :
Σ𝜖jk = (P.T @ Σy @ P)[:2, :2]

Pjk = P[:, :2]

Σy_hat = Pjk @ Σ𝜖jk @ Pjk.T
print('Σy_hat = \n', Σy_hat)

Σy_hat =
[[1.05 1.05 1.05 1.05 1.05 0.   0.   0.   0.   0.  ]
[1.05 1.05 1.05 1.05 1.05 0.   0.   0.   0.   0.  ]
[1.05 1.05 1.05 1.05 1.05 0.   0.   0.   0.   0.  ]
[1.05 1.05 1.05 1.05 1.05 0.   0.   0.   0.   0.  ]
[1.05 1.05 1.05 1.05 1.05 0.   0.   0.   0.   0.  ]
[0.   0.   0.   0.   0.   1.05 1.05 1.05 1.05 1.05]
[0.   0.   0.   0.   0.   1.05 1.05 1.05 1.05 1.05]
[0.   0.   0.   0.   0.   1.05 1.05 1.05 1.05 1.05]
[0.   0.   0.   0.   0.   1.05 1.05 1.05 1.05 1.05]
[0.   0.   0.   0.   0.   1.05 1.05 1.05 1.05 1.05]]


## Stochastic Difference Equation¶

Consider the stochastic second-order linear difference equation

$$y_{t} = \alpha_{0} + \alpha_{1} y_{y-1} + \alpha_{2} y_{t-2} + u_{t}$$

where $u_{t} \sim N \left(0, \sigma_{u}^{2}\right)$ and

$$\left[\begin{array}{c} y_{-1}\\ y_{0} \end{array}\right]\sim N\left(\mu_{\tilde{y}},\Sigma_{\tilde{y}}\right)$$

It can be written as a stacked system

$$\underset{\equiv A}{\underbrace{\left[\begin{array}{cccccccc} 1 & 0 & 0 & 0 & \cdots & 0 & 0 & 0\\ -\alpha_{1} & 1 & 0 & 0 & \cdots & 0 & 0 & 0\\ -\alpha_{2} & -\alpha_{1} & 1 & 0 & \cdots & 0 & 0 & 0\\ 0 & -\alpha_{2} & -\alpha_{1} & 1 & \cdots & 0 & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \cdots & \vdots & \vdots & \vdots\\ 0 & 0 & 0 & 0 & \cdots & -\alpha_{2} & -\alpha_{1} & 1 \end{array}\right]}}\left[\begin{array}{c} y_{1}\\ y_{2}\\ y_{3}\\ y_{4}\\ \vdots\\ y_{T} \end{array}\right]=\underset{\equiv b}{\underbrace{\left[\begin{array}{c} \alpha_{0}+\alpha_{1}y_{0}+\alpha_{2}y_{-1}\\ \alpha_{0}+\alpha_{2}y_{0}\\ \alpha_{0}\\ \alpha_{0}\\ \vdots\\ \alpha_{0} \end{array}\right]}}$$

We can compute $y$ by solving the system

$$y = A^{-1} \left(b + u\right)$$

We have

\mu_{y} = A^{-1} \mu_{b} \\ \begin{aligned} \Sigma_{y} &= A^{-1} E \left[\left(b - \mu_{b} + u \right) \left(b - \mu_{b} + u \right)^{\prime}\right] \left(A^{-1}\right)^{\prime} \\ &= A^{-1} \left(\Sigma_{b} + \Sigma_{u} \right) \left(A^{-1}\right)^{\prime} \end{aligned}

where

$$\mu_{b}=\left[\begin{array}{c} \alpha_{0}+\alpha_{1}\mu_{y_{0}}+\alpha_{2}\mu_{y_{-1}}\\ \alpha_{0}+\alpha_{2}\mu_{y_{0}}\\ \alpha_{0}\\ \vdots\\ \alpha_{0} \end{array}\right]$$$$\Sigma_{b}=\left[\begin{array}{cc} C\Sigma_{\tilde{y}}C^{\prime} & \boldsymbol{0}_{N-2\times N-2}\\ \boldsymbol{0}_{N-2\times2} & \boldsymbol{0}_{N-2\times N-2} \end{array}\right],\quad C=\left[\begin{array}{cc} \alpha_{2} & \alpha_{1}\\ 0 & \alpha_{2} \end{array}\right]$$$$\Sigma_{u}=\left[\begin{array}{cccc} \sigma_{u}^{2} & 0 & \cdots & 0\\ 0 & \sigma_{u}^{2} & \cdots & 0\\ \vdots & \vdots & \vdots & \vdots\\ 0 & 0 & \cdots & \sigma_{u}^{2} \end{array}\right]$$
In :
# set parameters
T = 80
T = 160
# coefficients of the second order difference equation
𝛼0 = 10
𝛼1 = 1.53
𝛼2 = -.9

# variance of u
𝜎u = 1.
𝜎u = 10.

# distribution of y_{-1} and y_{0}
𝜇y_tilde = np.array([1., 0.5])
Σy_tilde = np.array([[2., 1.], [1., 0.5]])

In :
# construct A and A^{\prime}
A = np.zeros((T, T))

for i in range(T):
A[i, i] = 1

if i-1 >= 0:
A[i, i-1] = -𝛼1

if i-2 >= 0:
A[i, i-2] = -𝛼2

A_inv = np.linalg.inv(A)

In :
# compute the mean vectors of b and y
𝜇b = np.ones(T) * 𝛼0
𝜇b += 𝛼1 * 𝜇y_tilde + 𝛼2 * 𝜇y_tilde
𝜇b += 𝛼2 * 𝜇y_tilde

𝜇y = A_inv @ 𝜇b

In :
# compute the covariance matrices of b and y
Σu = np.eye(T) * 𝜎u ** 2

Σb = np.zeros((T, T))

C = np.array([[𝛼2, 𝛼1], [0, 𝛼2]])
Σb[:2, :2] = C @ Σy_tilde @ C.T

Σy = A_inv @ (Σb + Σu) @ A_inv.T


## Application to Stock Price Model¶

Let

$$p_{t} = \sum_{j=0}^{T-t} \beta^{j} y_{t+j}$$

Form

$$\underset{\equiv p}{\underbrace{\left[\begin{array}{c} p_{1}\\ p_{2}\\ p_{3}\\ \vdots\\ p_{T} \end{array}\right]}}=\underset{\equiv B}{\underbrace{\left[\begin{array}{ccccc} 1 & \beta & \beta^{2} & \cdots & \beta^{T-1}\\ 0 & 1 & \beta & \cdots & \beta^{T-2}\\ 0 & 0 & 1 & \cdots & \beta^{T-3}\\ \vdots & \vdots & \vdots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & 1 \end{array}\right]}}\left[\begin{array}{c} y_{1}\\ y_{2}\\ y_{3}\\ \vdots\\ y_{T} \end{array}\right]$$

we have

$$\mu_{p} = B \mu_{y} \\ \Sigma_{p} = B \Sigma_{y} B^{\prime}$$
In :
𝛽 = .96

In :
# construct B
B = np.zeros((T, T))

for i in range(T):
B[i, i:] = 𝛽 ** np.arange(0, T-i)


Denote

$$z=\left[\begin{array}{c} y\\ p \end{array}\right]=\underset{\equiv D}{\underbrace{\left[\begin{array}{c} I\\ B \end{array}\right]}} y$$

Thus, $\{y_t\}_{t=1}^{T}$ and $\{p_t\}_{t=1}^{T}$ jointly follow the multivariate normal distribution $N \left(\mu_{z}, \Sigma_{z}\right)$, where

$$\mu_{z}=D\mu_{y}$$$$\Sigma_{z}=D\Sigma_{y}D^{\prime}$$
In :
D = np.vstack([np.eye(T), B])

In :
𝜇z = D @ 𝜇y
Σz = D @ Σy @ D.T


We can simulate paths of $y_{t}$ and $p_{t}$ and compute the conditional mean $E \left[p_{t} \mid y_{t-1}, y_{t}\right]$ using the MultivariateNormal class.

In :
z = np.random.multivariate_normal(𝜇z, Σz)
y, p = z[:T], z[T:]

In :
cond_Ep = np.empty(T-1)

sub_𝜇 = np.empty(3)
sub_Σ = np.empty((3, 3))
for t in range(2, T+1):
sub_𝜇[:] = 𝜇z[[t-2, t-1, T-1+t]]
sub_Σ[:, :] = Σz[[t-2, t-1, T-1+t], :][:, [t-2, t-1, T-1+t]]

multi_normal = MultivariateNormal(sub_𝜇, sub_Σ)
multi_normal.partition(2)

cond_Ep[t-2] = multi_normal.cond_dist(1, y[t-2:t])

In :
plt.plot(range(1, T), y[1:], label='$y_{t}$')
plt.plot(range(1, T), y[:-1], label='$y_{t-1}$')
plt.plot(range(1, T), p[1:], label='$p_{t}$')
plt.plot(range(1, T), cond_Ep, label='$Ep_{t}|y_{t}, y_{t-1}$')

plt.xlabel('t')
plt.legend(loc=1)
plt.show() In the above graph, the green line is what the price of the stock would be if people had perfect foresight about the path of dividends while the green line is the conditional expectation $E p_t | y_t, y_{t-1}$, which is what the price would be if people did not have perfect foresight but were optimally predicting future dividends on the basis of the information $y_t, y_{t-1}$ at time $t$.

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