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Let $X_1\sim \text{Unif}(0,1)$, $e\sim N(0,1)$ and $X_2=X_1+e$.

Is the joint distribution of $(X_1,X_2)$ multivariate Normal? Why or why not?

My solution:

By definition of a Multivariate normal distribution (in my textbook), a random vector $X\in \mathbb{R}^p$ is said to follow a multivariate normal distribution if $$X=\mu+AZ,$$

for some non-random $\mu\in\mathbb{R}^p$, some non-random $p\times l$ matrix $A$, and $Z=(Z_1,...,Z_l)^T$, where $Z_1,...,Z_1 \sim N(0,1)\text{(iid)},l\geq 1$

Now, assume $X_1$ and $e$ and uncorrelated. Then $\text{Var}(X_2)=\text{Var}(X_1)+\text{Var}(e)+2\text{Cov}(X_1,e)=\frac{1}{12}+1+0=\frac{13}{12}$

I can construct

$$A= \begin{pmatrix} \frac{1}{\sqrt{12}} & 0 \\ 0 & \sqrt{\frac{13}{12}} \end{pmatrix} $$

So that $$(X_1,X_2)=\mu + AZ$$

Where $\mu=(1,1)^T$, since $E[X_1]=\frac{1}{1-0}=1$ and $E[X_2]=E[X_1]+E[e]=1$ and $Z=(Z_1,Z_2)^T$, where $Z_1,Z_2\sim N(0,1)$

Assume now that $X_1$ and $e$ are correlated. Then $\text{Var}(X_2)=\frac{13}{12}+\text{Cov}(X_1,e)$. Let $c=\text{Var}(X_2)$

Then I can construct

I can construct

$$A= \begin{pmatrix} \frac{1}{\sqrt{12}} & 0 \\ 0 & \sqrt{c} \end{pmatrix} $$

Hence, by definition, the distribution is multivariate normal.

Is this solution wrong? If yes, could you please tell me exactly where I went wrong? I would appreciate that a lot

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  • $\begingroup$ If $X_1$ and $X_2$ were jointly normal, then marginally each would be normally distributed, which they're not. What makes you think they would be? $\endgroup$ May 15, 2021 at 13:28
  • $\begingroup$ @Aruralreader But I can't see what's exactly wrong with my solution either and that's why i'm asking in the first place. $\endgroup$
    – user926287
    May 15, 2021 at 13:35

1 Answer 1

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Joint distribution of $(X_1, X_2)$ can't be gaussian, since it's support is not equal to $\mathbb{R}^2$. For example, $$\mathbb{P}\left\{ X_1 \leq 0, X_2 \leq 0 \right\} = \mathbb{P}\left\{ X_1 \leq 0, X_1 + e \leq 0 \right\} \leq \mathbb{P}\left\{ X_1 \leq 0 \right\} = 0$$ Set $A=\left\{(x,y) \in \mathbb{R}^2 : x \leq 0, y\leq 0\right\}$ certainly has non-zero Lebesgue measure, so if $(X_1, X_2)$ was gaussian, $\mathbb{P}\left\{ (X_1, X_2) \in A\right\}$ would be non-zero.

EDIT: as for your prood, you state that $(X_1,X_2)^T =\mu + AZ$ for $$ \mu=(1,1)^T, A= \begin{pmatrix} \frac{1}{\sqrt{12}} & 0 \\ 0 & \sqrt{\frac{13}{12}} \end{pmatrix}, Z = (Z_1, Z_2)^T, Z_1,Z_2\sim N(0,1) $$ So, for $(Y_1, Y_2)^T = \mu + AZ = \left(1+\frac{1}{\sqrt{12}}Z_1, 1+\sqrt{\frac{13}{12}}Z_2\right)$, $Y_1 \sim N(1, \frac{1}{2})$ and $Y_2 \sim N(1, \frac{13}{12})$. Obviously, distribution of $Y_1$ is not equal to distribution of $X_1$, which is $\mathrm{Unif}(0, 1)$, and also, $Y_1$ and $Y_2$ are independent (equivalently, uncorrelated), but $cov(X_1, X_2) = cov(X_1, X_1) + cov(X_1, e) > 0$.

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  • $\begingroup$ Could you please clarify how does the example show that there is no support for $X_1$ at the interval $(-\infty,0)$? Also, I don't see exactly why $P(X_1\leq 0, X_1+e\leq 0)\leq P(X_1\leq 0)$. Otherwise I see why it's not multivariate normal and thank you for your answer $\endgroup$
    – user926287
    May 15, 2021 at 14:03
  • $\begingroup$ Let $A = \left\{ X_1 \leq 0\right\}$ and $B = \left\{ X_2 \leq 0\right\}$. $\mathrm{P}(A \cap B) \leq \mathrm{P}(A)$, since $A\cap B \subset A$. $\mathbb{P}\left\{ X_1 \leq 0\right\} = 0$, since $X_1 \sim \mathrm{Unif}(0,1)$. $\endgroup$
    – Yalikesi
    May 15, 2021 at 14:06
  • $\begingroup$ thank you very much for helping! $\endgroup$
    – user926287
    May 15, 2021 at 14:10

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