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I know how to prove the statement when two r.v. are independent. But how to prove it when the r.v. are correlated?

Edit:or counter example?

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    $\begingroup$ How can one prove what isn't true? (If $X$ and $Y$ have joint normal distribution, but are possibly not independent, then the sum is normal) $\endgroup$ Nov 12, 2013 at 3:13

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In some cases in which they are correlated it is not true that their sum has a normal (or "Gaussian") distribution. For example, suppose $X\sim N(0,1)$ and let $Y = \pm X$, where $+$ or $-$ is chosen randomly, and you pick "$+$" with probability $2/3$ and "$-$" with probability $1/3$, independently of the value of $X$. Then it's not hard to show that $Y\sim N(0,1)$, and $X$ is positively correlated with $Y$. But $\Pr(X+Y=0) = 1/3$, so $X+Y$ cannot be normally distributed.

If $X$ and $Y$ are jointly normally distributed, then it is indeed true that $X+Y$ is normally distributed regardless of the correlation between $X$ and $Y$. That's a trivial conseequence of a statement often taken to be the definition of joint normality: $(X,Y)$ has a bivariate normal distribution (so $X$ and $Y$ are "jointly normally distributed") precisely if the distribution of the pair $(X,Y)$ is such that every linear combination $aX+bY$, where $a,b$ are not random, is normally distributed.

Another definition of joint normality says there are standard normal random variables $Z_1,\ldots,Z_k$ such that $X$ is some linear combination of them, and $Y$ is some other linear combination of them. Then $X+Y$ is normally distributed because it's a linear combination of indendent normals $Z_1,\ldots,Z_k$.

Another characterization says the joint density is $$ \text{constant}\cdot\exp\left( \frac{-1}{2} [x-\mu,y-\nu]\begin{bmatrix} a & b \\ b & c \end{bmatrix}^{-1}\begin{bmatrix} x-\mu \\ y-\nu \end{bmatrix} \right) $$ where the matrix is a positive-definite. This is not quite equivalent to the foregoing to characterizations because it does not accomodate the case where the correlation is either $1$ or $-1$, or, in the case of a random vector with more than two components, the case where the matrix that is the variance is singular. Proving $X+Y$ is normally distributed in this case is something I would do by showing that if $\mathbf X\in\mathbb R^n$ and $\mathbf X\sim N_n(\mu,A)$, and $M\in\mathbb R^{k\times n}$ is a constant (i.e. not random) matrix, then $M\mathbf X \sim N_k(M\mu,MAM^\top)$. Maybe I'll add more about that later.

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In general, for jointly continuous random variables $X$ and $Y$ with joint pdf $f_{X,Y}(u,v)$,$$f_{X+Y}(\alpha) = \int_{-\infty}^\infty f_{X,Y}(u, \alpha-u)\,\mathrm du$$ which reduces to the convolution of the marginal densities $f_X$ and $f_Y$ when $X$ and $Y$ are independent random variables. For the special case when $X$ and $Y$ are jointly Gaussian random variables, the integral above can be computed by "completing the square" in the exponent so that you can pull out some things not dependent on $u$ outside the integral and leave a Gaussian density inside the integral, which then integrates to $1$. I assume that you know what a bivariate joint Gaussian density is.

If $X \sim N(\mu_X, \sigma_X^2)$ and $Y \sim N(\mu_Y, \sigma_Y^2)$ are jointly Gaussian random variables with correlation coefficient $\rho$, then $$X+Y \sim N(\mu_X+\mu_Y,\sigma_X^2+\sigma_Y^2+2\rho\sigma_X\sigma_Y)$$

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