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X1, and X2 are independent with the pdf: $f_{X}(x)= \sqrt{\frac{1}{2\pi x}e^{-x/2}}$ defined for x>0


What is the pdf of Y?

This is what I did so far:

$$f_{Y}(y) = \int_{0}^{\infty } f_{X_{1}}(y-x_{2})f_{X_{2}}(x_{2})dx_{2}$$ $$ = \int_{0}^{\infty } \sqrt{\frac{1}{2\pi (y-x_{2})}e^{-(y-x_{2})/2}} \sqrt{\frac{1}{2\pi x_{2}}e^{-x_{2}/2}} dx_{2}$$

$$=\int_{0}^{\infty } \sqrt{\frac{1}{2\pi (y-x_{2})}e^{(-y+x_{2})/2}} \sqrt{\frac{1}{2\pi x_{2}}e^{-x_{2}/2}} dx_{2}$$

$$=\int_{0}^{\infty } \sqrt{\frac{1}{2\pi (y-x_{2})}e^{(-y+x_{2})/2} \frac{1}{2\pi x_{2}}e^{-x_{2}/2}} dx_{2}$$

$$=\int_{0}^{\infty } \sqrt{\frac{1}{2\pi (y-x_{2})} \frac{1}{2\pi x_{2}}e^{-y/2}} dx_{2}$$

$$=\frac{1}{2\pi} \sqrt{e^{-y/2}} \int_{0}^{\infty } \sqrt{\frac{1}{ (y-x_{2})x_{2}} } dx_{2}$$

$$=\frac{1}{2\pi} \sqrt{e^{-y/2}} \int_{0}^{\infty } \sqrt{\frac{1}{ yx_{2}-(x_{2})^{2}} } dx_{2}$$

At this point I wasn't sure how to proceed with the integral, so I went to wolfram alpha. According to it, this integral does not converge, so not sure what to do now.

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Effectively $x_2$ only goes to $y$, not to $\infty$. (The density is $0$ at negatives.) – André Nicolas Feb 26 '13 at 9:08
In that case the integral evaluates to pi. Thanks, I see my mistake now. – kyphos Feb 26 '13 at 9:19
I must be making a stupid mistake somewhere, but I get the pdf to simplify to $$\frac{1}{\sqrt{2\pi x}}e^{-x/4}~~\text{for}~x> 0,$$ which is not quite the same as the Gamma pdf with parameters $(t,\lambda)=\left(\frac{1}{2},\frac{1}{4}\right)$ which is $$\frac{1}{\Gamma(t)}\lambda(\lambda x)^{t-1}e^{-\lambda x}=\frac{1}{\Gamma\left(\frac{1}{2}\right)}\frac{1}{4}\left(\frac{x}{4}\right)^{‌​-1/2}e^{-x/4} =\frac{1}{2\sqrt{\pi x}}e^{-x/4}~\text{for}~x>0.$$So, is there a typo in the problem? – Dilip Sarwate Feb 26 '13 at 13:24
"In that case the integral evaluates to $\pi$." But, then, your density works out to be $\frac{1}{2}e^{-y/4}$ which is not a valid pdf: its integral is $2$. – Dilip Sarwate Feb 26 '13 at 13:34
You are right. There was a typo, the pdf should be $$f_{X}(x) = \sqrt { \frac{1}{2 \pi x} } e^{-y/2}$$ That means the last line of what I have above should be: $$=\frac{1}{2\pi} e^{-y/2} \int_{0}^{y } \sqrt{\frac{1}{ yx_{2}-(x_{2})^{2}} } dx_{2}$$ And you would proceed forth from there same as Andre mentioned in the answer below. – kyphos Feb 27 '13 at 20:40
up vote 1 down vote accepted

Effectively $x_2$ only goes to $y$, not to $\infty$. (The density is $0$ at negatives.)

For the work that remains, complete the square. After making the right substitution, you should end up needing to find $\int\frac{du}{\sqrt{1-u^2}}$, which is $\arcsin u +C$.

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I am wondering if there is a typo in the given pdf? At first glance, I thought it is the sum of independent Gamma random variables with parameters $(1/2,1/4)$ which would give an exponential random variable with parameter $1/4$ as the answer, but the given pdf differs from the Gamma density by a constant multiplicative factor. – Dilip Sarwate Feb 26 '13 at 13:30

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