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Let $X$ and $Y$ have joint pdf $f(x,y) = 4e^{-2(x+y)}$ for $0 < x < \infty$, $0 < y < \infty$, and zero otherwise.

(a) Find the CDF of $W = X + Y$

(b) Find the joint pdf of $U = X/Y$ and $V=X$

(c) Find the marginal pdf of $U$

Could someone show me the statistics behind setting up the integration? I can do the computation myself. So for instance, for (b). I will at least need the Jacobian, $$\begin{vmatrix} U_x & U_y\\ V_x & V_y \end{vmatrix} = \dfrac{-X}{Y^2}$$

Then subbing, I get $f(u,v)=4e^{2(\frac{v}{u}+v)}$

And for the marginal, I am not continuing until I am sure (b) is right otherwise I will waste my time doing unnecessary computation.

(a) $$\int_{0}^{w} \int_{0}^{w-x} 4e^{-2(x+y)}dydx$$

Thanks

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Currently there is a typo in the problem. The proposed joint density has infinite integral. I do not know what is intended, though $4e^{-2(x+y)}$ would be a valid joint density. –  André Nicolas Nov 27 '12 at 3:56
    
No the bounds are intended. I am reading through a similiar problem and I understand where you are coming from –  Hawk Nov 27 '12 at 4:00
1  
OK, but do read your density function carefully. In my display, it reads $4e^{2(x+y)}$. –  André Nicolas Nov 27 '12 at 4:03
    
It is missing a minius sign, you are correct thank you –  Hawk Nov 27 '12 at 4:08
    
Jacobians are not supposed to be used as plug-and-chug instruments as you have done. For what values of $(x,y)$ does your expression hold? –  Dilip Sarwate Nov 27 '12 at 4:11

1 Answer 1

up vote 0 down vote accepted

You may also notice that the joint pdf density split into a function of the form $f(x)g(y)$, meaning that they are two independent gaussian random variables. Then everything become much easy with convolutions and relatives ...

share|improve this answer
    
gaussian random variables? –  Dilip Sarwate Nov 27 '12 at 4:07

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