Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am stuck on a question with two parts :

For $$f(x,y) = \begin{cases} a^2 e^{-ay} & 0 \le x \le y, \\ 0 & \text{otherwise} \end{cases}$$

a) Compute the distribution function and density of $Z = X+Y$

b) Find the joint distribution function and the densty of $(Y, X+Y)$

I have been able to calculate part a) by integrating the function by taking limits for $x$ from $0$ to $z/2$ and for $y$ $x$ to $z-x$ and I get the correct solution.

For part b) I am applying the theorem that $$f(y_1,y_2) = f(x_1(y_1,y_2),x_2(y_1,y_2))|J(y_1,y_2)|$$ where $J$ is the Jacobian.

But I am unsure about the limits for the integral. How do I proceed?

share|improve this question
    
You need to understand that there are lots of details that you have left out (e.g. for what values of $x$ and $y$ is it true that $f(x,y)=a^2e^{-ay}$) and which are needed in helping you get to the answer. Also, this looks like homework and if so, please add the homework tag. Using LaTex to format your expressions would also help a lot. –  Dilip Sarwate Dec 3 '12 at 16:01
    
Filled out the details –  user669083 Dec 3 '12 at 16:42
    
(Y, Z) = (Y, X+Y) is a linear transformation of (X,Y), so checking where (X,Y) is allowed to find the region in question. But confused –  user669083 Dec 3 '12 at 18:21
    
In part b) there is no integration to be performed at all, so I don't know what it means when you say "I am unsure about the limits for the integral". But you do need to determine the region in which your formula applies. –  Dilip Sarwate Dec 3 '12 at 18:39
    
For calculating the density I will have to integrate over the area, but how to calculate the area. Please help –  user669083 Dec 3 '12 at 19:23

1 Answer 1

up vote 1 down vote accepted

Insert the indicator functions in the densities (as they should be) and everything will follow.

Here, $f_{X,Y}(x,y)=g(y)\cdot[0\leqslant x\leqslant y]$ for some function $g$ whose exact expression will be irrelevant, and $(z,t)=(y,x+y)$ hence the Jacobian is $1$ and $(x,y)=(t-z,z)$.

This yields $f_{X,Y}(x,y)=g(z)\cdot[0\leqslant t-z\leqslant z]$ hence $$ f_{Y,X+Y}(z,t)=g(z)\cdot[z\leqslant t\leqslant 2z]. $$ Note that if $z\leqslant t\leqslant 2z$ then $z\leqslant 2z$ hence $z\geqslant0$ and $f_{Y,X+Y}$ is also $$ f_{Y,X+Y}(z,t)=g(z)\cdot[z\geqslant0,t\geqslant0,z\leqslant t\leqslant 2z]. $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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