Multivariate Normal Bookkeeping

For random variables $X_1,\dots, X_n$ , denote $\mathbf X := (X_1,\dots, X_n)^{\mathrm T}$ and define

$\mathbb E[\mathbf X] = (\mathbb E[X_1], \dots, \mathbb E[X_n])^{\mathrm T}.$

Define the covariance matrix $\boldsymbol{ \Sigma }_{\mathbf X} \in \mathcal M_{n \times n}( \mathbb R )$ by

$\displaystyle (\boldsymbol{ \Sigma }_{\mathbf X})_{i,j} = \mathrm{Cov}(X_i, X_j).$

Note that $\boldsymbol{ \Sigma }_{\mathbf X} = \mathbb E[(\mathbf X - \mathbb E[\mathbf X]) (\mathbf X - \mathbb E[\mathbf X])^{\mathrm T}]$ .

Problem 1. For i.i.d. $Z_1,\dots, Z_n \sim \mathcal N(0, 1)$ , define $\mathbf Z := (Z_1,\dots, Z_n)^{\mathrm T}$ . Calculate the p.d.f. $f_{\mathbf Z}$ of $\mathbf Z$ and evaluate $\boldsymbol{ \Sigma }_{\mathbf Z}$ .

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Solution. For each $I$ , the p.d.f. $f_{Z_i}$ of $Z_i$ is

$\displaystyle f_{Z_i}(z_i) = \frac{1}{\sqrt{2\pi}} e^{-\frac{z_i^2}{2}}.$

Since $Z_1,\dots, Z_n$ are independent,

$\begin{aligned} f_{\mathbf Z}(\mathbf z) &= \prod_{i=1}^n f_{Z_i}(z_i) = \frac 1{ \sqrt{ (2\pi)^{n} } } e^{-\frac 12 \sum_{i=1}^n z_i^2} = \frac 1{ \sqrt{ (2\pi)^{n} } } e^{-\frac 12 \mathbf z^{\mathrm T} \mathbf z}.\end{aligned}$

Furthermore,

${ ( \boldsymbol{ \Sigma }_{\mathbf Z} )}_{i,j} = \mathrm{Cov}(Z_i, Z_j) = \mathbb I_{\{0\}} (i-j) = \mathbf I_{i,j},$

where $\mathbf I \equiv \mathbf I_n$ denotes the $n \times n$ identity matrix.

Problem 2. For any invertible matrix $\mathbf A \in \mathcal M_{n \times n}(\mathbb R)$ and $\boldsymbol{\mu} \in \mathbb R^n$ , define $\mathbf X = \boldsymbol{\mu} + \mathbf A \mathbf Z$ . Calculate the p.d.f. $f_{\mathbf X}$ of $\mathbf X$ and evaluate $\boldsymbol{ \Sigma }_{\mathbf X}$ .

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Solution. We first assume the case $\boldsymbol{ \mu } = \mathbf 0$ . Define the bijective and continuous map. We observe that for $K \in \frak{B}(\mathbb R^n)$ ,

$\mathbb P_{\mathbf X}(K) = \mathbb P(\mathbf X \in K) = \mathbb P(\mathbf Z \in \mathbf A^{-1}(K)) = (\mathbb P_{\mathbf Z} \circ \mathbf A^{-1})(K).$

That is, $\mathbb P_{\mathbf X} = \mathbb P_{\mathbf Z} \circ \mathbf A^{-1}$ is the pushforward measure of $\mathbb P_{\mathbf Z}$ under $g$ . Letting $\lambda$ denote the $n$ -dimensional Lebesgue measure, we leave it as an exercise in linear algebra to verify that

$\displaystyle \frac{ \mathrm d\lambda(\mathbf A^{-1}\mathbf x) }{ \mathrm d\lambda(\mathbf x) } = |\mathbf A^{-1}| = \frac{1}{|\mathbf A|}.$

By a change of variables, for any $K \in \frak{B}(\mathbb R^n)$ ,

$\begin{aligned} \mathbb P_{\mathbf X} (K) &= \int_{\mathbb R^n} \mathbb I_K\, \mathrm d\mathbb P_{\mathbf X} \\ &= \int_{\mathbb R^n} \mathbb I_K\, \mathrm d\mathbb P_{\mathbf X} \\ &= \int_{\mathbb R^n} \mathbb I_K \circ \mathbf A \, \mathrm d\mathbb P_{\mathbf Z} \\ &= \int_{\mathbb R^n} \mathbb I_K( \mathbf A\mathbf z ) \cdot f_{\mathbf Z}(\mathbf z) \, \mathrm d\lambda(\mathbf z) \\ &= \int_{\mathbb R^n} \mathbb I_K(\mathbf x) \cdot f_{\mathbf Z}(\mathbf A^{-1} \mathbf x ) \, \mathrm d\lambda(\mathbf A^{-1} \mathbf x ) \\ &= \int_{\mathbb R^n} \mathbb I_K(\mathbf x) \cdot f_{\mathbf Z}(\mathbf A^{-1} \mathbf x ) \cdot \frac{1}{|\mathbf A|}\, \mathrm d\lambda(\mathbf x) \\ &= \int_{K} f_{\mathbf Z}(\mathbf A^{-1} \mathbf x ) \cdot \frac{1}{|\mathbf A|}\, \mathrm d\lambda(\mathbf x).\end{aligned}$

By the Radon-Nikodým theorem,

$\displaystyle \int_K f_{\mathbf X}(\mathbf x)\, \mathrm d\lambda(\mathbf x) = \mathbb P_{\mathbf X} (K) = \int_{K} f_{\mathbf Z}(\mathbf A^{-1} \mathbf x ) \cdot \frac{1}{|\mathbf A|}\, \mathrm d\lambda(\mathbf x),$

so that

$\begin{aligned} f_{\mathbf X}(\mathbf x) &= \frac{1}{|\mathbf A|} \cdot f_{\mathbf Z}(\mathbf A^{-1}\mathbf x) \\ &= \frac 1{ |\mathbf A| \cdot \sqrt{ (2\pi)^{n} } } \cdot e^{-\frac 12 \mathbf x^{\mathrm T} (\mathbf A\mathbf A^{\mathrm T})^{-1} \mathbf x } \\ &= \frac 1{ \sqrt{ (2\pi)^{n} \cdot |\mathbf A \mathbf A^{\mathrm T}| } } \cdot e^{-\frac 12 \mathbf x^{\mathrm T} (\mathbf A\mathbf A^{\mathrm T})^{-1} \mathbf x }.\end{aligned}$

Furthermore,

$\begin{aligned} \boldsymbol{\Sigma}_{\mathbf X} &= \mathbb E[(\mathbf A \mathbf Z - \mathbb E[\mathbf A \mathbf Z]) (\mathbf A \mathbf Z - \mathbb E[\mathbf A \mathbf Z])^{\mathrm T}] \\ &= \mathbb E[\mathbf A (\mathbf Z - \mathbb E[\mathbf Z]) (\mathbf Z - \mathbb E[\mathbf Z])^{\mathrm T}\mathbf A^{\mathrm T} ] \\ &= \mathbf A \mathbb E[(\mathbf Z - \mathbb E[\mathbf Z]) (\mathbf Z - \mathbb E[\mathbf Z])^{\mathrm T} ]\mathbf A^{\mathrm T} \\ &= \mathbf A \boldsymbol{\Sigma}_{\mathbf Z} \mathbf A^{\mathrm T} = \mathbf A \mathbf A^{\mathrm T} . \end{aligned}$

In the marginally more general case, $\mathbf X - \boldsymbol{\mu} = \mathbf A \mathbf Z$ . We leave it as an exercise to verify that $\mathbb E[\mathbf X] = \boldsymbol{ \mu }$ and $\boldsymbol{ \Sigma }_{\mathbf X} = \boldsymbol{ \Sigma }_{\mathbf A \mathbf Z} = \mathbf A\mathbf A^{\mathrm T}$ , so that finally

$\begin{aligned} f_{\mathbf X}(\mathbf x) = f_{\mathbf A \mathbf Z}(\mathbf x - \boldsymbol{\mu}) &= \frac 1{ \sqrt{ (2\pi)^{n} \cdot |\boldsymbol{ \Sigma }_{\mathbf A \mathbf Z}| } } \cdot e^{-\frac 12 (\mathbf x - \boldsymbol{\mu})^{\mathrm T} \boldsymbol{ \Sigma }_{\mathbf A \mathbf Z}^{-1} (\mathbf x - \boldsymbol{\mu}) }. \end{aligned}$

Definition 1. A random variable $\mathbf X$ is multivariate normal, denoted $\mathbf X \sim \mathcal N(\boldsymbol{ \mu }, \boldsymbol{ \Sigma })$ , if it has a p.d.f. $f_{\mathbf X}$ given by

$\displaystyle f_{\mathbf X}(\mathbf x) = \frac{1}{ \sqrt{ (2\pi)^n \cdot |\boldsymbol{ \Sigma } | } } \cdot e^{-\frac 12 (\mathbf x - \boldsymbol{\mu})^{\mathrm T} \boldsymbol{\Sigma}^{-1} (\mathbf x - \boldsymbol{\mu})}.$

In this case, we say that the mean of $\mathbf X$ is $\mathbb E[ \mathbf X ] = \boldsymbol{\mu}$ , and the covariance of $\mathbf X$ is $\boldsymbol{\Sigma}_{\mathbf{X}} = \boldsymbol{\Sigma}$ . For example, $\mathbf Z \sim \mathcal N(\mathbf 0, \mathbf I_n)$ .

Problem 3. Fix $\mathbf X \sim \mathcal N(\boldsymbol{ \mu }, \boldsymbol{\Sigma} )$ .

Prove that if $\boldsymbol{\Sigma}$ is diagonal whose diagonal entires are positive $\sigma_i^2 > 0$ , then $X_1,\dots, X_n$ are independent with $X_i \sim \mathcal N( \mu_i, \sigma_i^2 )$ .
Prove that for any invertible $\mathbf A \in \mathcal M_{n \times n}(\mathbb R)$ and $\boldsymbol{ \nu } \in \mathbb R^n$ , $\mathbf Y := \boldsymbol{\nu} + \mathbf A \mathbf X \sim \mathcal N(\boldsymbol{ \mu }_{\mathbf Y}, \boldsymbol{ \Sigma }_{\mathbf Y})$ for constants $\boldsymbol{ \mu }_{\mathbf Y}, \boldsymbol{ \Sigma }_{\mathbf Y}$ to be calculated.

(Click for Solution)

Solution. For the first point, if $\boldsymbol{\Sigma}$ is diagonal with $\boldsymbol{ \Sigma }_{i,i} = \sigma_i^2 > 0$ , then $\boldsymbol{\Sigma}^{-1}$ is diagonal with $\boldsymbol{ \Sigma }_{i,i} = 1/\sigma_i^2 > 0$ , and

$\displaystyle (\mathbf x - \boldsymbol{\mu})^{\mathrm T} \boldsymbol{\Sigma}^{-1} (\mathbf x - \boldsymbol{\mu}) = \sum_{i=1}^n \frac{ (x_i - \mu_i)^2 }{ \sigma_i^2 }.$

Some algebra then yields

$\begin{aligned} f_{\mathbf X}(\mathbf x) &= \frac{1}{ \sqrt{ (2\pi)^n \cdot |\boldsymbol{ \Sigma } | } } \cdot e^{-\frac 12 (\mathbf x - \boldsymbol{\mu})^{\mathrm T} \boldsymbol{\Sigma}^{-1} (\mathbf x - \boldsymbol{\mu})} \\ &= \prod_{i=1}^n \frac{1}{\sigma_i \sqrt{ 2\pi } } \cdot e^{-\frac 12 \sum_{i=1}^n \frac{ (x_i - \mu_i)^2 }{ \sigma_i^2 } } \\ &= \prod_{i=1}^n \frac{1}{\sigma_i \sqrt{ 2\pi } } \cdot \prod_{i=1}^n e^{-\frac{ (x_i - \mu_i)^2 }{ 2\sigma_i^2 } } \\ &= \prod_{i=1}^n \frac{1}{\sigma_i \sqrt{ 2\pi } } e^{-\frac{ (x_i - \mu_i)^2 }{ 2\sigma_i^2 } } = \prod_{i=1}^n f_{X_i}(x_i). \end{aligned}$

Therefore, $X_1,\dots, X_n$ are independent with $X_i \sim \mathcal N(\mu_i , \sigma_i^2)$ .

For the second point, follow the proof in Problem 2 with needful bookkeeping to obtain

$\begin{aligned} f_{\mathbf Y}(\mathbf y) &= f_{\mathbf A\mathbf X}(\mathbf y - \boldsymbol{\nu}) = \frac 1{|\mathbf A|} \cdot f_{\mathbf X}(\mathbf A^{-1}(\mathbf y - \boldsymbol{\nu})), \end{aligned}$

which simplifies to Definition 1 with $\mathbb E[\mathbf Y] = \boldsymbol{\nu} + \mathbf A \boldsymbol{\mu}$ and $\boldsymbol{\Sigma}_{\mathbf Y} = \mathbf A \boldsymbol{\Sigma}_{\mathbf X} \mathbf A^{\mathrm T}$ .

Problem 4. Let $\mathbf X \sim \mathcal N(\boldsymbol{ \mu } , \boldsymbol{ \Sigma })$ . Prove that there exists an invertible matrix $\mathbf A \in \mathcal M_{n \times n}(\mathbb R)$ such that $\mathbf X = \boldsymbol{ \mu } + \mathbf A \mathbf Z$ . In particular, each $X_i$ is normally distributed.

(Click for Solution)

Solution. By definition,

We first assume $\boldsymbol{\mu} = \mathbf 0$ . If we can find some invertible matrix $\mathbf B$ such that $\mathbf B \mathbf X = \mathbf Z$ , then we can set $\mathbf A := \mathbf B^{-1}$ . To that end, we note that $\boldsymbol{\Sigma}$ is symmetric since

$\displaystyle \boldsymbol{\Sigma}_{i,j} = \mathrm{Cov}(X_i, X_j) = \mathrm{Cov}(X_j, X_i) = \boldsymbol{\Sigma}_{j,i} > 0.$

By the spectral theorems, $\boldsymbol{\Sigma}$ is orthogonally diagonalisable. That is, there exists an orthogonal matrix $\mathbf P$ (i.e. $\mathbf P^{-1} = \mathbf P^{\mathrm T}$ ) and a diagonal matrix $\mathbf D$ such that

$\mathbf{P}^{\mathrm T} \boldsymbol{\Sigma} \mathbf{P} = \mathbf{D}\quad \Rightarrow \quad \boldsymbol{\Sigma} = \mathbf{P} \mathbf{D} \mathbf{P}^{\mathrm T}.$

Furthermore, the entries in the diagonal of $\mathbf D$ are all positive, denoted $\sigma_i^2 > 0$ . Define $\boldsymbol{\Delta}$ by

$\displaystyle \boldsymbol{\Delta}_{i,j} := \begin{cases} \sigma_i, & i = j, \\ 0, & i \neq j. \end{cases}$

Then $\boldsymbol{\Delta}\boldsymbol{\Delta}^{\mathrm T} = \mathbf{D}$ , so that

$\boldsymbol{\Sigma} = \mathbf{P} \mathbf{D} \mathbf{P}^{\mathrm T} = \mathbf{P} \boldsymbol{\Delta}\boldsymbol{\Delta}^{\mathrm T} \mathbf{P}^{\mathrm T} = (\mathbf{P} \boldsymbol{\Delta}) (\mathbf{P} \boldsymbol{\Delta})^{\mathrm T}.$

Now define $\mathbf B := ( \mathbf{P} \boldsymbol{\Delta})^{-1}$ . By Problem 3, $\mathbf B \mathbf X$ is multivariate normal. We calculate its expectation via

$\mathbb E[\mathbf B \mathbf X] = \mathbf B \mathbf E[\mathbf X] = \mathbf B \mathbf 0 = \mathbf 0$

and its covariance matrix via

$\begin{aligned}\boldsymbol{\Sigma_{\mathbf B \mathbf X}} &= \mathbb E[(\mathbf B \mathbf X)^{\mathrm T} (\mathbf B \mathbf X)] \\ &= \mathbf B \mathbb E[\mathbf X^{\mathrm T} \mathbf X] \mathbf B^{\mathrm T} \\ &= ( \mathbf{P} \boldsymbol{\Delta})^{-1} (\mathbf{P} \boldsymbol{\Delta}) (\mathbf{P} \boldsymbol{\Delta})^{\mathrm T} (( \mathbf{P} \boldsymbol{\Delta})^{-1})^{\mathrm T} = \mathbf I. \end{aligned}$

Therefore, $\mathbf B \mathbf X \sim \mathcal N(\mathbf 0, \mathbf I)$ . Since $\mathbf Z \sim \mathcal N(\mathbf 0, \mathbf I)$ , we conclude $\mathbf B \mathbf X = \mathbf Z$ , so that $\mathbf X = \mathbf B^{-1}\mathbf Z = \mathbf A \mathbf Z$ , as required.

For the general case, $\mathbf X - \boldsymbol{\mu}$ has expectation $\mathbf 0$ , and so by the first result, there exists some invertible matrix $\mathbf A$ such that

$\mathbf X - \boldsymbol{\mu} = \mathbf A \mathbf Z \quad \Rightarrow \quad \mathbf X = \boldsymbol{\mu} + \mathbf A \mathbf Z,$

as required.

Problem 5. Define $W_1 := \bar X_n$ and for each $i > 1$ , define $W_i := X_i - \bar X_n$ . If $X_1,\dots,X_n \sim \mathcal N(\mu, \sigma^2)$ are i.i.d., prove that $\bar X_n$ and $(W_2,\dots,W_n)$ are independent. Deduce that $\bar X_n$ and $S_n^2$ defined by

$\displaystyle S_n^2 := \frac 1{n-1} \sum_{i=1}^n (X_i - \bar X_n)^2$

are independent.

(Click for Solution)

Solution. For each $i > 1$ ,

$\displaystyle W_i = \sum_{j=1}^n \underbrace{ \left( \mathbb I_{\{ i \}}(j) -\frac{ 1 }{n} \right) }_{\mathbf A_{i,j}} X_j =: \sum_{j=1}^n \mathbf A_{i, j}X_j.$

Furthermore, define $\mathbf A_{1, j} = 1/n$ . Defining $\mathbf W := (W_1,\dots,W_n)^{\mathrm T}$ , $\mathbf W = \mathbf A \mathbf X$ . If $X_1,\dots,X_n$ are i.i.d. normally distributed with variance $\sigma^2$ , then by the converse of the first point in Problem 3 (which holds), $\mathbf X$ is multivariate normal. By the second point in Problem 3, $\mathbf W$ is multivariate normal.

We turn to evaluate its covariance matrix. Firstly, to shorten computations,

$\begin{aligned} \mathrm{Cov}(X_i, \bar X_n) &= \frac 1n \cdot \sum_{k=1}^n \mathrm{Cov}(X_i, X_k) \\ &= \frac 1n \cdot \mathrm{Cov}(X_i, X_i) = \frac{\sigma^2}{n}. \end{aligned}$

Furthermore,

$\begin{aligned} \mathrm{Cov}(\bar X_n, \bar X_n) &= \frac 1n \cdot \sum_{k=1}^n \mathrm{Cov}(X_i, \bar X_n) \\ &= \frac 1n \cdot \sum_{k=1}^n \frac{\sigma^2}{n} = \frac 1n \cdot n \cdot \frac{\sigma^2}n = \frac{\sigma^2}n. \end{aligned}$

Thus, for $j > 1$ ,

$\begin{aligned} (\boldsymbol{\Sigma}_{\mathbf W})_{1,j} &= \mathrm{Cov}(W_1, W_j) \\ &= \mathrm{Cov}(\bar X_n, X_j - \bar X_n) \\ &= \mathrm{Cov}(X_j , \bar X_n) - \mathrm{Cov}(\bar X_n,\bar X_n) \\ &= \frac{\sigma^2}{n} - \frac{\sigma^2}{n} = 0.\end{aligned}$

Therefore, using properties involving the exponential function,

$f_{\mathbf W}(\mathbf w) = f_{W_1}(w_1) \cdot f_{W_2,\dots,W_n}(w_2,\dots,w_n).$

Hence, $W_1$ and $(W_2,\dots,W_n)$ are independent. For the second claim, we first note that

$\displaystyle 0 = \sum_{i=1}^n X_i - n\bar X_n = \sum_{i=1}^n (X_i - \bar X_n) = X_1 - \bar X_n + \sum_{i=2}^n (X_i - \bar X_n),$

so that $X_1 - \bar X_n = -\sum_{i=2}^n (X_i - \bar X_n) = -\sum_{i=2}^n W_i$ . It follows that

$\begin{aligned} S_n^2 &= \frac 1{n-1} \left( (X_1 - \bar X_n)^2 + \sum_{i=2}^n (X_i - \bar X_n)^2 \right) \\ &= \frac 1{n-1} \left( (X_1 - \bar X_n)^2 + \sum_{i=2}^n W_i^2 \right) \\ &= \frac 1{n-1} \left( \left( \sum_{i=2}^n W_i \right)^2 + \sum_{i=2}^n W_i^2 \right). \end{aligned}$

Since $S_n^2$ can be written purely in terms of $(W_2,\dots, W_n)$ , it is independent of $W_1 = \bar X_n$ .

Problem 6. Using the definitions and notation in Problem 5, prove that for any $n \in \mathbb N_{>1}$ ,

$\displaystyle (n-1) \cdot S_n^2 = (n-2) \cdot S_{n-1}^2 + \frac{n-1}{n} \cdot (X_n - \bar X_{n-1})^2.$

Deduce that there exist i.i.d. $Z_1,\dots, Z_{n-1} \sim \mathcal N(0, 1)$ such that

$\displaystyle \frac{(n-1) \cdot S_n^2}{\sigma^2} = \sum_{i=1}^{n-1} Z_i^2.$

(Click for Solution)

Solution. We first observe that

$\begin{aligned} \bar X_n - \bar X_{n-1} &= \frac{1}{n} \cdot \sum_{i=1}^{n-1} X_i + \frac{1}{n} \cdot X_n - \frac 1{n-1} \cdot \sum_{i=1}^{n-1} X_i \\ &= -\frac{1}{n(n-1)} \cdot \sum_{i=1}^{n-1} X_i + \frac{1}{n} \cdot X_n = \frac 1n \cdot ( X_n - \bar X_{n-1} ). \end{aligned}$

Performing algebruh,

$\begin{aligned} &(n-1) \cdot S_n^2 \\ &= \sum_{i=1}^n \left( X_i - \bar X_{n-1} - \frac 1n \cdot ( X_n - \bar X_{n-1})\right)^2 \\ &= \sum_{i=1}^n \left[ (X_i - \bar X_{n-1})^2 - 2 \cdot (X_i - \bar X_{n-1}) \cdot \frac 1n \cdot ( X_n - \bar X_{n-1}) + \frac 1{n^2} \cdot ( X_n - \bar X_{n-1})^2 \right] \\ &= (n-2) \cdot S_{n-1}^2 + (X_n - \bar X_{n-1})^2 - \frac{2}{n} \cdot ( X_n - \bar X_{n-1})^2 + \frac 1{n} \cdot ( X_n - \bar X_{n-1})^2 \\ &= (n-2) \cdot S_{n-1}^2 + \frac{n-1}{n} \cdot (X_n - \bar X_{n-1})^2. \end{aligned}$

We now prove the final equation by induction. In the case $n = 2$ , we have

$\begin{aligned} \frac{S_2^2}{\sigma^2} &= \frac 1{\sigma^2} \cdot \frac 12 \cdot ((X_1 -\bar X_1)^2 + (X_2 -\bar X_1)^2) = \frac{(X_2 - X_1)^2}{2\sigma^2}. \end{aligned}$

Then $Z_1 := (X_2 - X_1)/(\sigma \sqrt 2) \sim \mathcal N(0, 1)$ , so that $S_2^2/\sigma^2 = Z_1^2$ , as required. For the induction step, suppose that the statement holds for $n = k$ . Then

$\displaystyle k \cdot S_{k+1}^2 = (k-1) \cdot S_k^2 + \frac{k}{k+1} \cdot (X_{k+1} - \bar X_k)^2.$

By relabelling the result in Problem 5, we have that $S_k^2$ is independent of $\bar X_k$ and $X_{k+1}$ . By the induction hypothesis, there exist i.i.d. $Z_1,\dots, Z_{k-1} \sim \mathcal N(0, 1)$ such that

$\displaystyle \frac{ (k-1) \cdot S_k^2 }{\sigma^2} := \sum_{i=1}^{k-1} Z_i^2.$

Furthermore, $X_{k+1} - \bar X_k \sim \mathcal N(0, \sigma^2 \cdot (k+1)/k)$ so that

$\displaystyle Z_{k+1} := \frac{\sqrt{k}}{\sqrt{k+1}} \cdot \frac{ X_{k+1} - \bar X_k }{\sigma} = \frac{X_{k+1} - \bar X_k}{ \sigma \cdot \sqrt{ \frac{k+1}{k} } } \sim \mathcal N(0, 1).$

Therefore,

$\displaystyle \displaystyle \frac{ k \cdot S_{k+1}^2 }{ \sigma^2 } = \frac{ (k-1) \cdot S_{k}^2 }{ \sigma^2 } + \frac{k}{k+1} \cdot \frac{ (X_{k+1} - \bar X_{k})^2 }{ \sigma^2 } = \sum_{i=1}^{k-1} Z_i^2 + Z_k^2 = \sum_{i=1}^{k} Z_i^2,$

as required.

—Joel Kindiak, 25 Jul 25, 2018H

KindiakMath

Multivariate Normal Bookkeeping

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Multivariate Normal Bookkeeping

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