Suppose two random variables X and Y have joint pdf $f(x,y) = 3x^2e^{-y} $ when 0 < x < 1, y > 0. f(x,y) = 0 otherwise.
a. Find the marginal pdf of X
b. Find the conditional density of Y given X and evaluate P(Y > 1|X = 0.5).
c. Show that Cov(X,Y) = 0. Are X and Y independent?
d. Compute E[2X(3X + Y)].
I've compute the marginal pdf of X to be 3x^2 when 0 < x < 1. Then that means the conditional density of Y given X is $\frac{f(x,y)}{f_x(x)} = \frac{3x^2e^{-y}}{(3x^2)}$. I'm having trouble calculating P(Y > 1|X = 0.5). Would it be correct to calculate 1 - P(Y <= 1|X = 0.5)? And would I just integrate $\frac{f(x,y)}{f_x(x)}$ from 0 to 1?
For part C I know that Cov(X,Y) = E(XY) - E(X)E(Y). Would E(XY) = $\int^ \infty_0\int^1_0(xy*f(x,y))dxdy$? Have I chosen my bounds correctly?
You have $f_{\small X,Y}(x,y)= 3 x^2\mathrm e^{-y}\mathbf 1_{x\in(0;1)}\mathbf 1_{y\in(0;\infty)}$.
So indeed:
(a) $f_{\small X}(x)=3x^2\mathbf 1_{x\in(0;1)}\require{cancel}\cancelto{1}{~\int_0^1\mathrm e^{-y}\mathrm d y~}$
(b) $f_{\small Y\mid X}(y\mid x)=e^{-y}\mathbf 1_{y\in(0;\infty)}\mathbf 1_{x\in(0;1)}$ and hence:$$\begin{align}\mathsf P(Y\gt 1\mid X=0.5) ~&=~\int_1^\infty \mathrm e^{-y}\,\mathrm d y\\[1ex]&=~1-\int_0^1\mathrm e ^{-y}\,\mathrm d y\end{align}$$
(c) $\mathsf E(XY)=\int_0^1\int_0^\infty xy\cdot 3x^2\mathrm e^{-y}\,\mathrm d y\,\mathrm d x$ and hence $$\begin{align}\mathsf E(XY)&=\int_0^1 x\cdot 3x^2\,\mathrm d x\cdot\int_0^\infty y\cdot \mathrm e^{-y}\,\mathrm d y\\[2ex]&=\text{hmm...}\end{align}$$
Well done.
(d) So you are on track to evaluate $\mathsf E(2X(3X+Y))$.