# Is the joint distribution of $(X_1,X_2)$ multivariate Normal?

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

• 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? May 15, 2021 at 13:28
• @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.
– user926287
May 15, 2021 at 13:35

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 $$cov(X_1, X_2) = cov(X_1, X_1) + cov(X_1, e) > 0$$.
• 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
• 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 $X_1 \sim \mathrm{Unif}(0,1)$. May 15, 2021 at 14:06