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Transformation of f(x,y) = 1 to f(z) where Z=XY

  1. Oct 9, 2013 #1
    This is from a chapter on distributions of two random variables. Let X and Y have the pdf f(x,y) = 1, 0<x<1 and 0<y<1, zero elsewhere. Find the cdf and pdf of the product Z=XY.

    My current approach has been to plug in X=Z/Y in the cdf P(X<=x) , thus P(Z/Y<=x), and integrate over all values of Y. The integral I've tried to solve is ∫(0 to 1)∫(0 to Z/Y) 1 dxdy (sorry for the formatting).

    This is somehow wrong, because from that integral I get z*(ln(1) - ln(0)), which is undefined (unless I'm solving the integral incorrectly, which would actually be a relief).

    The correct answer is -ln(z), 0<z<1, though it's amiguous as to whether that's the cdf or pdf.

    Thanks for any solutions or suggestions about where my approach is going wrong.
     
  2. jcsd
  3. Oct 9, 2013 #2

    mathman

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    -ln(z) is a density function.

    When doing the integral note that the upper limit on x is min(1,z/y).
     
  4. Oct 9, 2013 #3
    Another way of seeing the same thing... [It looks a little less direct, but there's value in thinking about sums of exponential random variables.]

    Let [itex]\tilde X = -ln(X)[/itex], [itex]\tilde Y = -ln(Y)[/itex], and [itex]\tilde Z = -ln(Z) = \tilde X+\tilde Y[/itex], all taking values on [itex](0,\infty)[/itex]. Define the CDF [itex]G(\tilde x)= P(\tilde X\leq \tilde x)= P(X\geq e^{-\tilde x})=1-e^{-\tilde x}[/itex], which is of course also the CDF of [itex]\tilde Y[/itex]. Differentiating, we get the PDF [itex]g(\tilde x)=e^{-\tilde x}[/itex]. Now we can compute [tex]P(\tilde Z> \tilde z\geq \tilde X) = \int_0^{\tilde z} P(\tilde x+\tilde Y> \tilde z) g(\tilde x) \text{d}\tilde x = \int_0^{\tilde z} [1- G( \tilde z-\tilde x)] g(\tilde x) \text{d}\tilde x = \int_0^{\tilde z} e^{\tilde x-\tilde z} e^{-\tilde x} \text{d}\tilde x = \int_0^{\tilde z} e^{-\tilde z} \text{d}\tilde x = \tilde z e^{-\tilde z}.[/tex] This yields [tex]P(Z<z\leq X) = P(\tilde Z> -ln(z)\geq \tilde X) = [-ln(z)]z.[/tex] Finally, we notice that (as [itex]Z\leq X[/itex]) [tex]P(Z<z)=P(X<z)+P(Z<z\leq X) = z - [-ln(z)]z = [1-ln(z)]z.[/tex] If your end goal is a density, you can differentiate for a PDF of [itex]-ln(z)[/itex].
     
  5. Oct 10, 2013 #4
    Thanks for the responses. Before I dive into economicsnerd's solution (wow), I'm a bit confused by min(1,Z/Y). When Z<Y, min(1,Z/Y) is Z/Y. When Z>Y, it's 1. So...

    ∫(0 to 1)∫(0 to Z/Y)dxdy where 0<z<y<1
    ∫(0 to z)∫(0 to 1)dxdy where 0<y<z<1

    Isn't the first integral still undefined? Or am I applying this incorrectly?
     
  6. Oct 10, 2013 #5
    The CDF of [itex]Z[/itex] is [tex]F(z)= P(Z\leq z) = 1- P(Z>z) = 1-P(X\geq Z > z) = 1-\int_z^1 \int_{\frac{z}{x}}^1 1\enspace \text{d}y \enspace \text{d}x = 1-\int_z^1 \left(1-\frac{z}{x}\right) \enspace \text{d}x.[/tex] What do you get when you compute that integral?
     
  7. Oct 10, 2013 #6
    I get your solution. Awesome.

    I'm confused as to why P(Z>z) = P(X>=Z>z). Are we able to say this since 0<x<1, 0<y<1, so any multiplication of x and y is less than x and y? I just never would've seen that.

    I'm also stumped about the construction of this integral. The outer integral z to 1 dx is fine, but why do we choose z/x as the lower limit of the integral over dy? When we plug in z/x for y in the equation 0<y<1, we get 0<z/x<1... and from this it's still not clear to me that z/x is the lower limit.

    edit: it's probably because we flipped P(Z<z) to P(Z>z)...

    Thanks for your help!
     
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