# Obtaining marginal distributions from the bivariate normal

Let $(X, Y)$ have a normal distribution with mean $(\mu_X, \mu_Y)$, variance $(\sigma_X^2, \sigma_Y^2)$ and correlation $\rho$. I want to know the corresponding marginal densities.

All I found so far was the well-known density expressions for $X\sim N(\mu_X, \sigma_X^2)$ and $Y\sim N(\mu_Y, \sigma_Y^2)$, but isn't that just for $X \perp Y$? Shouldn't $\rho$ appear in the expressions? How do I calculate the marginals of any $X$ and $Y$? I think using the definition will end up in an integral that cannot be solved analytically…

• The essence of the problem can be appreciated by inspecting the PDF for $((X-\mu_x)/\sigma_X,(Y-\mu_Y)/\sigma_Y)$. The logarithm of the part that depends on $X$ and $Y$ looks like $-\frac{1}{2}(X^2 + Y^2 - 2XY\rho)/(1-\rho^2)$. Viewing $X$ as a constant for the purpose of integrating out $Y$, it is evident that you must compute a Normal integral, which is easy and has an exact solution. What's left depends only on $X$ and $\rho$: by definition, it's the marginal distribution.
– whuber
Mar 1, 2013 at 23:12
• @Dilip I definitely did mean to include $\rho$ in that sentence. The joint PDF is bivariate normal but it's correlated. When you complete the square you will introduce a factor of $\exp(\frac{1}{2}\rho^2/(1-\rho^2))$ and you will be integrating the exponential of $-\frac{1}{2}(Y-2X\rho)^2/(1-\rho^2)$ wrt $Y$. There is a question of whether the $\rho$ disappears after all the smoke clears :-).
– whuber
Mar 1, 2013 at 23:40
• It's not just for $X\perp Y.$ The correlation $\rho$ will appear in the joint and conditional distributions, but not in the marginal distributions. Dec 31, 2017 at 14:45

Following up on whuber's comment, \begin{align} \left.\left.\frac{1}{2(1-\rho^2)}\right(x^2+y^2-2\rho xy\right) &=\left.\left.\frac{1}{2(1-\rho^2)}\right(x^2-\rho^2x^2+(y^2-2\rho xy + \rho^2x^2)\right)\\ &= \frac{x^2}{2} + \frac{1}{2}\left(\frac{y-\rho x}{\sqrt{1-\rho^2}}\right)^2 \end{align} and so $$\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dy = \frac{e^{-x^2/2}}{\sqrt{2\pi}} \int_{-\infty}^\infty \frac{e^{-(y-\rho x)^2/2(1-\rho^2)}}{\sqrt{1-\rho^2}\sqrt{2\pi}}\,\mathrm dy.$$ I will leave it to the OP to complete the details and determine whether $\rho$ disappears or not when the integral is evaluated.
• I understand you forced the means to equal $0$ and the variances $1$ for simplicity. Let $u\triangleq y-\rho x$. Then $du = dy$, and by defining $\mu_U\triangleq 0$ and $\sigma_U\triangleq\sqrt{1-\rho^2}$, it's clear that the integral has to be $1$. That is, $\rho$ disappears indeed!
• @quirik Which second expression do you mean? (If you right-click on any expression or formula, it will show you the LaTEx commands that produced it. Copy it and paste it (surrounded by \$signs) in your response) Jul 16, 2016 at 20:20 • @Dilip Sarwate I am confused with$\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dy = \frac{e^{-x^2/2}}{\sqrt{2\pi}} \int_{-\infty}^\infty \frac{e^{-(y-\rho x)^2/2(1-\rho^2)}}{\sqrt{1-\rho^2}\sqrt{2\pi}}\,\mathrm dy$. Since am I self-studying this topic, I would be very gratefull if you could show me the whole procedure. Jul 16, 2016 at 20:23 • @quirik Replace$f_{X,Y}(x,y)$by the exact expression for the bivariate normal density of standard normal variables with correlation coefficient$\rho$. Compare the argument of$\exp(\,\,)\$ on the left side with the the calculations immediately above. Can you see how the right side integral follows? Jul 16, 2016 at 21:31