Probability Density Function

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!

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Originally Posted by Darkprince

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!

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Here is an outline with many details you will need to fill in:

$\displaystyle \displaystyle cdf = F(y) = \Pr(Y \leq y) = \Pr(\sqrt{X} \leq y) = \Pr(X \leq y^2) = ....$

Therefore $\displaystyle \displaystyle pdf = f(y) = \frac{dF}{dy} = ....$

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

FY(y)=P(Y<=y)
=P(Sqrt(X)<= y)
=P(X<=y^2)
= 1-e^(-y^2)

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!

Kind regards!

looks ok, but in the last line it's y's not x's, i.e., y>0

Thank you very much Beagle! Have a nice day!
Kind regards :)