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1. ## Transformation question

I would appreciate if anyone could help with this question. It is from Probability and Statistics with Applications: A Problem Solving Text Asimow and Mawell
8-3
Suppose that the random variable X has an exponential distribution with mean one. Let the random variable Y= sqrt (X)

(a) Find the probability density function for Y.
(b) Find the expected value for Y.

For part (a) I got the density function for Y: f(y)= (density function for X)*(dx/dy) = e^-x * 2y= 2ye^(-y^2)

Using that answer I attempted to solve part (b) by doing the integral from 0 to infinity of y*f(y). But I cannot figure out a solution using integration tables or integration by parts. I suspect I've made an error in the density function of Y.

2. Originally Posted by bhopkins
I would appreciate if anyone could help with this question. It is from Probability and Statistics with Applications: A Problem Solving Text Asimow and Mawell
8-3
Suppose that the random variable X has an exponential distribution with mean one. Let the random variable Y= sqrt (X)

(a) Find the probability density function for Y.
(b) Find the expected value for Y.

For part (a) I got the density function for Y: f(y)= (density function for X)*(dx/dy) = e^-x * 2y= 2ye^(-y^2)

Using that answer I attempted to solve part (b) by doing the integral from 0 to infinity of y*f(y). But I cannot figure out a solution using integration tables or integration by parts. I suspect I've made an error in the density function of Y.
Is the answer \sqrt{\pi}/4? If so you might have to get it in form of the standard normal and then see which factor is missing and use the symmetry of standard normal random variable.

3. Thank you for replying. I finally figured it out sorry I didn't update the thread.

the answer is sqrt(pi)/2 I had to use an improper integral that converges I found in an earlier chapter it was the integral from negative infinity to positive infinity or e^(-(z/2)^2) is equal to sqrt(2 pi). Or half of that since the integral starts at zero. I integrated sqrt(x)*f(x) instead of the transformation y*f(y) and did a u-substitution. . . then integration by parts.

As it turns out after reading more in that previous chapter, there is a property of gamma distribution, The integral from zero to infinity or x^(a-1)*e^(-(1/b)x) is equal to b^(a) * gamma(a) . which would have be a lot quicker.
There was another integral I came across that would have helped me if I tried to integrate y*f(y) but I can't find it. I will go back and look at your way just to see, but the key was finding a improper integral with e in it that equaled some thing with pi in it. I was looking at a calculus book for the longest but i guess it was a probability specific problem.

I very much appreciate your help. It seems you have no more exams, so I'll try to pass the kindness on by help someone else if I am able.