If X has an exponential distribution with mean 1, find the probability density function of Y=Sqrt(X). I know that c.d.f of X is 1-e^(-kχ) for x>=0 and 0 for x<0. Also its p.d.f is ke^(-kx) for x>-0 and 0 for x<0. By giving us that mean = 1 we know that 1/k (the expectance) equals to 1 so k=1. But then how do i find the p.d.f of y=sqrt(X)?? After a bit of searching I found something about transformations of random variables but I am not sure. If someone can give me this exercise's solution with baby steps I would appreciate it much!
Thanks in advance! Kind regards!
Thanks for the reply! To be sure that i got the right answer, this is my complete work and please let me know if it is complete and correct!
FX(x)=1-e^(-x) for x>=0
0 for x<0
Hence p.d.f = dFY(y)/dy
= 2y*e(-y^2) for x>=0 and o for x<0
Thanks in advance and thanks again for the reply!