# Find marginal distribution of $Y$ where $Y\mid X$ is $N(a_1+a_2X,\sigma_1^2)$ and $X$ is $N(\mu,\sigma^2)$?

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!


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.