Consider $X=(X_1,\ldots,X_n)^T\sim\mathcal{N}(\mu,V)$. Show that then $X_i\sim\mathcal{N}(\mu_i,V_{ii})$ for all $1\leqslant i\leqslant n$.

Good day!

Ok, I have to determine the marginal distribution. To do so I have to calculate

$$f_i(X_i)=\int_{-\infty}^{\infty}\ldots\int_{-\infty}^{\infty}f(x_1,\ldots,x_n)d\, x_1\ldots d\, x_{i-1}d\, dx_{i+1}\ldots d\, x_n$$


$$f(x_1,\ldots,x_n)=(2\pi)^{-n/2}\text{det}(V)^{-1/2}\exp\left\{-\frac{1}{2}(x-\mu)^T V^{-1}(x-\mu)\right\}.$$

My problem is that I do not know how to calculate



$$-\frac{1}{2}(x-\mu)^T V^{-1}(x-\mu).$$

Anyhow, it is $$V=\begin{pmatrix}var(X_1) & \ldots & cov(X_1,X_n)\\\vdots & \ddots & \vdots\\cov(X_n,X_1) & \ldots & var(X_n)\end{pmatrix}=\begin{pmatrix}\sigma_1^2 & \ldots & cov(X_1,X_n)\\\vdots & \ddots & \vdots\\cov(X_n,X_1) & \ldots & \sigma_n^2\end{pmatrix}.$$

Can somebody help me, please?




Please do correct me if I am wrong. But to my knowledge there is a theorem saying that $X$ is multivariate normal distributed iff $a^TX$ is univariate normal distributed for any $a\in\mathbb{R}^n$.

So my suggestion is to choose $a=e_i=(0,\ldots,0,1,0,\ldots,0$ with the $1$ at the ith position. Then $a^TX=X_i$. So we know that $X_i$ is univariate normal distributed.

Because it is $E(X_i)=\mu_i$ and $Var(X_i)=\sigma_i^2=V_{ii}$ I think everything is shown?

As I said: Please correct me if I am wrong.


Ok, I have to determine the marginal distribution.

Actually no, because there are other ways and because this would restrict your solution to the case where $V$ is invertible although the result is more general.

Instead, note that $V$ is nonnegative symmetric hence $V=LL^T$ for some square matrix $L$ of size $n\times n$ and $$X=\mu+LZ,$$ where $Z$ is standard normal, that is, the distribution of $Z$ is normal $(0,I_n)$. Let $u_i$ denote the deterministic vector of size $n\times1$ with entries $0$ except the $i$th entry which is $1$. Then $$X_i=u_i^TX=u_i^T\mu+u_i^TLZ.$$ As a linear combination of the i.i.d. standard normal random variables $(Z_j)$, $u_i^TLZ$ is centered normal hence $X_i$ is normal with mean $u_i^T\mu=\mu_i$ and variance $$\sigma_i^2=E((u_i^TLZ)^2). $$ The size of $u_i^TLZ$ is $1\times1$ hence $u_i^TLZ=(u_i^TLZ)^T=Z^TL^Tu_i$ and an equivalent formula for the variance of $X_i$ is $$\sigma_i^2=E(u_i^TLZZ^TL^Tu_i)=u_i^TLE(ZZ^T)L^Tu_i.$$ By definition $E(ZZ^T)=I_n$ hence $\sigma_i^2=u_i^TLL^Tu_i=u_i^TVu_i=V_{ii}$, QED.

The last identity follows from the fact that for every matrix $M=(M_{ij})$ of size $n\times n$ and every $(i,j)$, $u_i^TMu_j=M_{ij}$.

Likewise, for every $i\ne j$, the covariance $C_{ij}$ between $X_i$ and $X_j$ is $$ C_{ij}=E((X_i-\mu_i)(X_j-\mu_j))=E((u_i^TLZ)(u_j^TLZ)), $$ and $u_j^TLZ=(u_j^TLZ)^T=Z^TL^Tu_j$ hence $$ C_{ij}=E(u_i^TLZZ^TL^Tu_j)=u_i^TLE(ZZ^T)L^Tu_j, $$ that is, $$ C_{ij}=u_i^TLL^Tu_j=u_i^TVu_j=V_{ij}. $$

  • $\begingroup$ Can I use the following theorem that I found in my script? If $X\sim N(\mu,V)$, then $U=V^{-1/2}(X-\mu)\sim N(0,I_n)$ which means that $U_i\sim N(0,1), i=1,...,n$ and the $U_i$ are independent. $\endgroup$
    – mathfemi
    May 21 '14 at 11:38
  • $\begingroup$ If $V$ is invertible and if one interprets suitably $V^{-1/2}$ (that is, if one replaces this imprecise notation by what I explain in my answer), yes. $\endgroup$
    – Did
    May 21 '14 at 11:53

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