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Transformation of Random Variables (Z = X-Y)

  1. Dec 29, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose we have a function, f(x,y) = e^-x * e^-y , 0<=x< ∞, 0<=y<∞,
    where X and Y are exponential random variables with mean = 1. (For those who may not know, all this means is ∫(x*e^(-x) dx) from 0 to ∞ = 1, and the same for y)

    Suppose we want to transform f(x,y) into f(z), where the transformation is Z = X-Y. Find f(z)

    2. Relevant equations
    f(x,y) = e^-x * e^-y , 0<=x< ∞, 0<=y<∞
    Z = X-Y

    3. The attempt at a solution

    So I decided to transform -Y into W. So we have -Y<=W, which implies Y>=-W
    ∫e^-y dy from -w to ∞...
    -e^-y from -w to ∞
    0 + e^-(-w)

    Differentiate wrt w

    f(w) = e^w -∞ < w <= 0

    So now we have Z = X+W
    Z-W = X
    We'll just let W stay as is for this problem.

    The jacobian of this transformation is 1.

    So we have f(z) = ∫e^-(z-w)*e^w dw from -∞ to z,

    This becomes ∫e^-z*e^2w dw from -∞ to z
    This becomes e^-z * (e^2w)/2 from -∞ to z
    This becomes e^-z * ((e^2z)-0)/2
    Which becomes (e^z)/2

    The domain of z is -∞ to ∞, however, this integral does not evaluate to 1. As a matter of fact, it does not even converge.

    Any help would be greatly appreciated!!
  2. jcsd
  3. Dec 30, 2011 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    Let f be the marginal density of X or Y. P{Z <= z} = int_{y=0..infinity} f(y)*P{X-Y <= z|Y=y} dy, and P{X-Y <= z|Y=y} = P{X <= y+z|Y=y} = P{X <= y+z} because X and Y are independent. You can get P{X <= y+z} and so do the integral. The final result is perfectly fine for all z in R and does, indeed, integrate to 1. However, you need to be careful about integration limits, since f(y) = exp(-y)*u(y) and F(y+z) = [1-exp(-y-z)]*u(y+z), where u(.) is the unit step function [u(t) = 0 for t < 0 and u(t) = 1 for t > 0]. BTW: please either use brackets (so write e^(2z)) or the "[ S U P ]" button (so write e2z) or else write "exp", as I have done. That way there is no chance of misreading what you type. Alternatively, you could use LaTeX.

  4. Jan 15, 2012 #3
    So sorry, I saw your solution a while ago and it really helped me. I just remembered that I never thanked you/gave you credit for your solution. How do I give you credit for helping me?
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