1
$\begingroup$

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)

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.