Probability Density Function

[Dark_Tranquilllity]

If X has an exponential distribution with mean 1, find the probability density function of Y=Sqrt(X). I don't even know how to handle this type of question, any help will be appreciated. Thanks in advance!

Kind regards!

 

[galactus]

The exponential distribution is given by \(\displaystyle \Large\frac{1}{\theta}e^{\frac{-x}{\theta}}\)

Where the mean for the exp. dist. is \(\displaystyle {\theta}\).

 

[Dark_Tranquilllity]

I know that c.d.f of X is 1-e^(-k?) 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)??