0
$\begingroup$

$X$ and $Y$ are random variables. The marginal distribution of $X$ is uniformly on $[0,1]$. The conditional distribution $f_{Y|X}$ is $U[x,x+1]$ for $0 \leq x \leq 1$.

Now derive the joint pdf of $X$ and $Y$ and the mean of $Y$.

I got that the joint pdf is simply $1$ if $0\leq x\leq 1$ and $x\leq y\leq x+1$ and for the mean of $Y$ I got $\dfrac{x}{2}$. But I am really unsure.

$\endgroup$
0
$\begingroup$

$$f_{X,Y}(x,y)=f_{Y|X}(y|x)f_X(x)=1, \, x\in[0,1],\, y \in[x, x+1]$$ $$\mathsf E(Y|X=x)=\int_{x}^{x+1}y.1dy=x+\frac{1}{2}\Rightarrow \mathsf E(Y|X)=X+\frac{1}{2}$$ $$\mathsf E Y=\mathsf E(\mathsf EY|X)=\mathsf E(X+\frac{1}{2})=\int_{0}^{1}(x+\frac{1}{2}).1dx=1$$

$\endgroup$

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.