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I'm just confused on how to set up and start this problem. I'm confident that once I start down the right path, I'll have little issue.

Let $p_1$ denote a bivariate normal distribution $N(0, 0, 1, 1, 0)$, and $p_2$ denote a bivariate normal distribution $N(0, 0, 1, 1, \rho)$. Suppose they have joint density:

$$p(x, y) = \frac{1}{2}p_1(x,y) + \frac{1}{2}p_2(x,y)$$.

I have to show that $X$ and $Y$ have marginally normal densities, but that their joint density is normal iff $\rho=0$.

I want to accomplish this through the use of matrix notation, instead of via the pdf.

The problem I'm having is that the problem doesn't say $p_1$ and $p_2$ are independent, and I believe they aren't since, for example:

$$Cov(X_{p_1},X_{p_2})=\mathbb{E}(X_{p_1}X_{p_2})+\mathbb{E}(X_{p_1})\mathbb{E}(X_{p_2})=\mathbb{E}(X^2)+0=1$$.

And the same for $Y$. Even though they're not independent, I guess I could still reduce them to their marginal distributions:

$$ p_1(X) \sim N\left(\begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} \mu_{X_1}\\\mu_{Y_1} \end{pmatrix}, \begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} \right) $$

And so on for $p_2(X)$. Then, I would find $p(X)$ by:

$$\mathbb{E}(p(X))= \frac{1}{2}\mathbb{E}(p_1(X))+\frac{1}{2}\mathbb{E}(p_2(X))$$

and

$$Var(p(X))=\left(\frac{1}{2}\right)^2 Var(p_1(X)) + \left(\frac{1}{2}\right)^2 Var(p_2(X)) + 2 Cov(X,X)$$

And similarly for Y.

Even if they were independent, though, I wouldn't know exactly how to set up the joint density part of the problem in matrix terms. For example, would it be (using $\mathbf{W}=\begin{pmatrix} 1/2 & 1/2 \end{pmatrix}$):

$$\mathbf{p}\sim N\left( \mathbf{W} \boldsymbol{\mu}_{p_1} + \mathbf{W} \boldsymbol{\mu}_{p_2}, \mathbf{W} \boldsymbol{\Sigma}_{p_1} \mathbf{W}^{T} + \mathbf{W} \boldsymbol{\Sigma}_{p_2} \mathbf{W}^{T} \right)$$

I would appreciate some insight into how to set up this problem.

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  • $\begingroup$ You lost me at the outset, because (a) "$\frac{1}{2}p_1(x,y)+\frac{1}{2}p_1(x,y)$" is just a complicated way to express $p_1(x,y),$ suggesting there's an error. Do you mean one of those subscripts to be "$2$"? If so, the results I proved at stats.stackexchange.com/a/431387/919 and stats.stackexchange.com/a/429877/919 easily give the answer: the first equates bivariate normality with normality of all linear combinations while the second will show that when $\rho\ne0,$ some linear combinations cannot be Normal (because they involve mixtures of Normals with different variances). $\endgroup$
    – whuber
    Oct 17, 2019 at 20:10
  • $\begingroup$ It is a typo! Sorry! And I hate to be a bother, but both of those responses use characteristic functions to prove the result, and I don't know anything about characteristic functions. Is there any other way to show this in terms of matrices or even the pdf? $\endgroup$
    – mdawgig
    Oct 17, 2019 at 20:23
  • $\begingroup$ That linear combinations of bivariate normal variables are Normal is easy to show using the PDF. The other result--about mixtures--can also be obtained using the PDF, but it takes a little more work. In your case, one way is to compute the kurtosis of a linear combination of the mixture ($X+Y$ will work) and show it's not that of a Normal distribution unless $\rho=0.$ $\endgroup$
    – whuber
    Oct 17, 2019 at 21:17
  • $\begingroup$ BTW, your notation is strange and might be misleading you. For examples of effective notation and computations with mixture densities, see stats.stackexchange.com/a/16609/919. Perhaps it would help to visualize these distributions: some are plotted in my answer at stats.stackexchange.com/a/326678/919. Note that in your case the two components are concentric. $\endgroup$
    – whuber
    Oct 17, 2019 at 21:43

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