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Math Help - Probability Density Function

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    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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    Quote Originally Posted by Darkprince View Post
    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!
    Here is an outline with many details you will need to fill in:

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

    Therefore \displaystyle pdf = f(y) = \frac{dF}{dy} = ....
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    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!
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  4. #4
    MHF Contributor matheagle's Avatar
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    looks ok, but in the last line it's y's not x's, i.e., y>0
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    Thank you very much Beagle! Have a nice day!
    Kind regards
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