# Joint PDF and Conditional Expectation

Let joint probability density function $X_1, X_2$ be

$$f_{1,2}(x_1, x_2) = 2e^{-x_1-x_2}I_{(0<x_1<x_2<\infty)}$$

When $Y_1 = X_1/X_2, Y_2 = X_2$,

(a) Derive PDF of $Y_1$

(b) Derive $E(Y_2 \mid Y_1)$ and $Var(Y_2 \mid Y_1)$

For the (a) I had used Jacobian Transformation and get joint pdf of $y_1$ and $y_2$ which is equal to $pdf(y_1,y_2) = 2y_2e^{-y_2(y_1+1)}I_{(0<y_1<1,\; 0<y<\infty)}$then, marginalize it about $y_1$ then I get $pdf(y_1) = \dfrac{2}{(y_1+1)^2}I_{(0<y_1<1)}$

However, for (b), to derive E(Y_2\mid Y_1), do I have to find $pdf(Y2\cap Y_1)/pdf(Y_1)$ then find expectation or any other approach is possible to shorten or simplify the procedure? Any hint? (Because to find $pdf(Y_2 \cap Y_1)$, I had checked these two variables are not independent..I have some problem to proceed)

For the first part note that the quotient distribution always has the form

$$f_Z(z) = \int_{-\infty}^{\infty} |y| f_{X_1, X_2}(zy, y) dy$$

(it is easy to see this by calculating $P(X_1 < z X_2)$ and then differentiating by $z$ to obtain the PDF).

In this case, this is

$$\int_{-\infty}^{\infty} |y| 2e^{-zy -y} I_{(0<zy < y <\infty)} dy.$$

Since $0 < y$, the integrand is non-zero for $z < 1$, and for that we have

$$\int_{0}^{\infty} y 2e^{-y(z + 1)} dy,$$

which is easy to solve.

For the second part, note that

$$E[Y_2|Y_1=y_1] = E[X_2|X_1 = y_1 X_2] = \int x_2 2e^{-y_1 x_2 -x_2}I_{(0<y_1 x_2 <x_2<\infty)} d x_2$$

which is very similar to the first part.