Covariance and expectation of joint continuous random variables

Question

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Firstly, for part (a) I have tried calculating this by integrating the conditional pdf along with y in the integral, but just get a messy integral and not sure how you could get a specific number from this. Is this the correct way to do it? Second, for part (b) again I tried calculating E(x) first but get a horrible integration and not sure how to find the pdf of Y... can anyone help??

Answer 1

Hints:

a)

For a fixed $x\neq0$ function $f_{Y\mid X}\left(y\mid x\right)=x^{2}e^{-yx^{2}}$ can be recognized as PDF of exponential distribution with parameter $\lambda=x^{2}$.

Consequently $\mathbb{E}\left[Y\mid X=x\right]=\frac{1}{\lambda}=x^{-2}$ so that $\mathbb{E}\left[Y\mid X\right]=X^{-2}$.

b)

• The PDF of $X$ betrays that there is symmetry wrt to $0$. Based on that it can be shown that for any function $u:\mathbb R\to\mathbb R$ such that $\mathbb Eu(X)$ exists and $u(-x)=-u(x)$ for every $x$ we have: $$\mathbb Eu(X)=\mathbb Eu(-X)=-\mathbb Eu(X)\text{ hence }\mathbb Eu(X)=0$$

• $\mathbb{E}XY=\mathbb{E}\left[\mathbb{E}\left[XY\mid X\right]\right]=\mathbb{E}\left[X\mathbb{E}\left[Y\mid X\right]\right]=\mathbb{E}\left[XX^{-2}\right]=\mathbb{E}X^{-1}$.