2
$\begingroup$

Let a random variable $X$ be normal $N(\mu,\sigma^2)$ and let the conditional distribution of $Y$ given $X$ be normal $N(a_1+a_2X,\sigma_1^2)$.

a)Find the joint probability density function of $X$ and $Y$.

b)Find the marginal distribution of $Y$ and the correlation coefficient of $X$ and $Y$.

For (a), I just multiplied the conditional density of $Y$ given $X$ and density of $X$; and I think it's ok. For (b), I tried to write their joint density in the form of bivariate normal but couldn't do that. On the other hand, we know that if the random variables $X$ and $Y$ are bivariate normal then, the conditional distribution of $X$ given $Y$ is normal with mean $E[X\mid Y]$ and variance $(1-\rho^2)\sigma_X^2$. But is that true that if $X$ is normal and $Y$ given $X$ is normal, then they are bivariate normal? So that I can do (b) easily, or is there another way to solve this question?

Thanks!

$\endgroup$
1
$\begingroup$

$\newcommand{\var}{\operatorname{var}}\newcommand{\cov}{\operatorname{cov}}$For part (a) you're ok except that I would do some routine simplications to make it clear that the two variables are in symmetrical roles.

You have $Y\mid X\sim N(a_1+a_2X,\sigma_1^2)$ and $X\sim N(\mu,\sigma^2)$.

From $Y\mid X\sim N(a_1+a_2 X,\sigma_1^2)$ we get $Y-(a_1+a_2 X)\mid X\sim \underbrace{\quad N(0,\sigma_1^2)\quad}_{\text{No ``$X$'' here.}}$.

If the conditional distribution of a random variable given $X$ does not depend on $X$, then we can conclude two things:

  • That random variable (in this case $Y-(a_1+x_2 X)$) is independent of $X$; and
  • That conditional distribution is also the marginal distribution (in this case the marginal distribution of $Y-(a_1+a_2 X)$). Consequently we have $\var(Y-(a_1+a_2 X))=\sigma_1^2$.

So $$ \overbrace{\var(Y) = \var(Y-(a_1 + a_2 X)) + \var(a_1+a_2 X)}^{\text{by independence}} = \sigma_1^2 + a_2^2\var(X) = \sigma_1^2+a_2^2\sigma^2. $$ And $$ \operatorname{E}(Y) = \operatorname{E}(\operatorname{E}(Y\mid X)) = \operatorname{E}(a_1+a_2 X) = a_1+a_2 \mu. $$ That gives us the marginal distribution of $Y$.

Next $$ 0 = \cov(Y-(a_1+a_2 X),X) = \cov(Y,X) - a_2\cov(X,X), $$ so $\cov(Y,X)=a_2\sigma^2$. Then use $\operatorname{corr}(X,Y)=\dfrac{\cov(X,Y)}{\sqrt{\var(X)\var(Y)}}$.

As for bivariate normality, consider linear combinations $cX+dY$. Show that that is a linear combination of the two independent random variables mentioned above.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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