8
$\begingroup$

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…

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ 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. $\endgroup$
    – whuber ♦
    Mar 1, 2013 at 23:12
  • $\begingroup$ @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 :-). $\endgroup$
    – whuber ♦
    Mar 1, 2013 at 23:40
  • 1
    $\begingroup$ 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. $\endgroup$
    – Michael Hardy
    Dec 31, 2017 at 14:45

1 Answer 1

Reset to default
8
$\begingroup$

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.

Improve this answer
$\endgroup$
6
  • 2
    $\begingroup$ 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! $\endgroup$
    – Luke
    Mar 2, 2013 at 4:08
  • $\begingroup$ @Dilip Sarwate Could you please provide how did you get the second expression? $\endgroup$
    – Quirik
    Jul 16, 2016 at 19:05
  • $\begingroup$ @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) $\endgroup$
    – Dilip Sarwate
    Jul 16, 2016 at 20:20
  • $\begingroup$ @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. $\endgroup$
    – Quirik
    Jul 16, 2016 at 20:23
  • $\begingroup$ @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? $\endgroup$
    – Dilip Sarwate
    Jul 16, 2016 at 21:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.