(Expectation)Consider the joint density $f(x,y)=c(x−y)e^{−x}, 0 \le y \le x$.

Question

Consider the joint density $f(x,y)=c(x−y)e^{−x}, 0 \le y \le x$.

a) Determine the value of c.

b) Calculate the marginal of Y.

c) Calculate the expectation E(Y).

a)$1=\int_0^x c(x−y)e^{−x} = \frac{cx^2e^{-x}}{2} = 1 \to c = \frac{2e^x}{x^2}, x \ne 0 $

b)$F_Y(Y) = \int_{-\infty}^{\infty} f(x,y)dx =\int_y^x\frac{\frac{2e^x}{x^2}(x^2-y)e^{-x}}{2}d_x = \frac{x^2 + y}{x}- y - 1$

c)$E[Y] = \int_{-\infty}^{\infty}yc(x−y)e^{−x}dy =\int_0^xy\frac{2e^x}{x^2}(x−y)e^{−x}dy =\frac{x}{3}$

This is what I managed to do, but I don't know if it's right, it's looking kind of weird, am I by the monos on the right track?

Thanks for any help.

Answer 1

$X$ and $Y$ are jointly distributed, so the total probability over the common support $0 \le Y \le X$ must be $1$. In other words, you cannot equate the constant $c$ to some function of $x$. That makes no sense: $c$ is a number that does not depend on $X$ or $Y$.

Specifically, we must find $c$ such that $$\int_{x=0}^\infty \int_{y=0}^x f_{X,Y}(x,y) \, dy \, dx = 1.$$ This is because the support of the density is the set of all ordered pairs $(X,Y)$ such that $0 \le Y \le X$.

For the second part, you would compute $$f_Y(y) = \int_{x=y}^\infty f_{X,Y}(x,y) \, dx.$$ This result must not be a function of $x$, only of $y$. Note the interval of integration is $x \in [y, \infty)$ because $0 \le y \le x$.

For the third part, you would compute $$\operatorname{E}[Y] = \int_{y=0}^\infty y f_Y(y) \, dy$$ using the result from the second part. Note that the interval of integration is not $y \in [0,x]$, because the marginal density of $Y$ does not depend on $x$.

Answer 2

I will just point out (a). What you have done for (a) is not correct. You are supposed to get a constant that does not depend on $x$ or $y$.

Write $A$ for the set of all $(x,y) \in \mathbb{R}^2$ such that $0 \leqslant y \leqslant x$. To find $c$, you need to apply the definition of density correctly, that is:

\begin{equation*} \int_Ac(x-y)e^{-x}dm(x,y)=1 \end{equation*}

Here $m$ denotes the Lebesgue measure in $\mathbb{R}^2$. Therefore:

\begin{equation*} c=(\int_A(x-y)e^{-x}dm(x,y))^{-1} \end{equation*}

This is legitimate as the integrand positive in the interior of $A$. All you need to do is to compute out the integral explicitly.