Let X and Y have the joint density f(x, y) = e −y , for 0 ≤ x ≤ y. (a) Find Cov(X, Y ) and the correlation of X and Y . (b) Find E(X|Y = y) and E(Y |X = x). (c) Find E(X) and Var(X) using, respectively, Theorems 1 and 2 in Note Set 8.

a. I assume the following definitions for covariance and correlation:[tex]\mathrm{Cov}[X,Y]=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y][/tex][tex]\mathrm{Corr}[X,Y]=\dfrac{\mathrm{Cov}[X,Y]}{\sqrt{\mathrm{Var}[X]\mathrm{Var}[Y]}}[/tex]Recall that[tex]E[g(X,Y)]=\displaystyle\iint_{\Bbb R^2}g(x,y)f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy[/tex]where [tex]f_{X,Y}[/tex] is the joint density, which allows us to easily compute the necessary expectations (a.k.a. first moments):[tex]E[XY]=\displaystyle\int_0^\infty\int_0^yxye^{-y}\,\mathrm dx\,\mathrm dy=3[/tex][tex]E[X]=\displaystyle\int_0^\infty\int_0^yxe^{-y}\,\mathrm dx\,\mathrm dy=1[/tex][tex]E[Y]=\displaystyle\int_0^\infty\int_0^yye^{-y}\,\mathrm dx=2[/tex]Also, recall that the variance of a random variable [tex]X[/tex] is defined by[tex]\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2[/tex]We use the previous fact to find the second moments:[tex]E[X^2]=\displaystyle\int_0^\infty\int_0^yx^2e^{-y}\,\mathrm dx\,\mathrm dy=2[/tex][tex]E[Y^2]=\displaystyle\int_0^\infty\int_0^yy^2e^{-y}\,\mathrm dx\,\mathrm dy=6[/tex]Then the variances are[tex]\mathrm{Var}[X]=2-1^2=1[/tex][tex]\mathrm{Var}[Y]=6-2^2=2[/tex]Putting everything together, we find the covariance to be[tex]\mathrm{Cov}[X,Y]=3-1\cdot2\implies\boxed{\mathrm{Cov}[X,Y]=1}[/tex]and the correlation to be[tex]\mathrm{Corr}[X,Y]=\dfrac1{\sqrt{1\cdot2}}\implies\boxed{\mathrm{Corr}[X,Y]=\dfrac1{\sqrt2}}[/tex]b. To find the conditional expectations, first find the conditional densities. Recall that[tex]f_{X,Y}=f_{X\mid Y}(x\mid y)f_Y(y)=f_{Y\mid X}(y\mid x)f_X(x)[/tex]where [tex]f_{X\mid Y}[/tex] is the conditional density of [tex]X[/tex] given [tex]Y[/tex], and [tex]f_X[/tex] is the marginal density of [tex]X[/tex].The law of total probability gives us a way to obtain the marginal densities:[tex]f_X(x)=\displaystyle\int_x^\infty e^{-y}\,\mathrm dy=\begin{cases}e^{-x}&\text{for }x\ge0\\0&\text{otherwise}\end{cases}[/tex][tex]f_Y(y)=\displaystyle\int_0^ye^{-y}\,\mathrm dx=\begin{cases}ye^{-y}&\text{for }y\ge0\\0&\text{otherwise}\end{cases}[/tex]Then it follows that the conditional densities are[tex]f_{X\mid Y}(x\mid y)=\begin{cases}\frac1y&\text{for }0\le x<y\\0&\text{otherwise}\end{cases}[/tex][tex]f_{Y\mid X}(y\mid x)=\begin{cases}e^{x-y}&\text{for }0\le x<y\\0&\text{otherwise}\end{cases}[/tex]Then the conditional expectations are[tex]E[X\mid Y=y]=\displaystyle\int_0^y\frac xy\,\mathrm dy\implies\boxed{E[X\mid Y=y]=\frac y2}[/tex][tex]E[Y\mid X=x]=\displaystyle\int_x^\infty ye^{x-y}\,\mathrm dy\implies\boxed{E[Y\mid X=x]=x+1}[/tex]c. I don't know which theorems are mentioned here, but it's probably safe to assume they are the laws of total expectation (LTE) and variance (LTV), which say[tex]E[X]=E[E[X\mid Y]][/tex][tex]\mathrm{Var}[X]=E[\mathrm{Var}[X\mid Y]]+\mathrm{Var}[E[X\mid Y]][/tex]We've found that [tex]E[X\mid Y]=\frac Y2[/tex] and [tex]E[Y\mid X]=X+1[/tex], so that by the LTE,[tex]E[X]=E[E[X\mid Y]]=E\left[\dfrac Y2\right]\implies E[Y]=2E[X][/tex][tex]E[Y]=E[E[Y\mid X]]=E[X+1]\implies E[Y]=E[X]+1[/tex][tex]\implies2E[X]=E[X]+1\implies\boxed{E[X]=1}[/tex]Next, we have[tex]\mathrm{Var}[X\mid Y]=E[X^2\mid Y]-E[X\mid Y]^2=\dfrac{Y^2}3-\left(\dfrac Y2\right)^2\implies\mathrm{Var}[X\mid Y]=\dfrac{Y^2}{12}[/tex]where the second moment is computed via[tex]E[X^2\mid Y=y]=\displaystyle\int_0^y\frac{x^2}y\,\mathrm dx=\frac{y^2}3[/tex]In turn, this gives[tex]E\left[\dfrac{Y^2}{12}\right]=\displaystyle\int_0^\infty\int_0^y\frac{y^2e^{-y}}{12}\,\mathrm dx\,\mathrm dy\implies E[\mathrm{Var}[X\mid Y]]=\frac12[/tex][tex]\mathrm{Var}[E[X\mid Y]]=\mathrm{Var}\left[\dfrac Y2\right]=\dfrac{\mathrm{Var}[Y]}4\implies\mathrm{Var}[E[X\mid Y]]=\dfrac12[/tex][tex]\implies\mathrm{Var}[X]=\dfrac12+\dfrac12\implies\boxed{\mathrm{Var}[X]=1}[/tex]