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Convolution math problem

  1. Mar 16, 2007 #1
    Given two independent variables with these simple density functions:

    [tex]f(x) = \left\lbrace \begin{array}{ll}
    \frac{1}{2} &\mbox{ if } 0 < x < 2 \\
    0 &\mbox{otherwise}
    \end{array} \right. [/tex]

    [tex]g(y) = \left\lbrace \begin{array}{ll}
    \frac{1}{3} &\mbox{ if } 1 < y < 4 \\
    0 &\mbox{otherwise}
    \end{array} \right. [/tex]

    if Z = X + Y, we can easily find the density function h(z) by the convolution of f and g, so for example
    [tex]\int_{-\infty}^{\infty}f(z-t)g(t)dt = \int_{0}^{2}f(z-t)\cdot \frac{1}{2}dt = \frac{1}{2} \int_{z-2}^{z}f(u)du = \frac{1}{6} \int_{z-2}^{z}du[/tex]

    and since g lives on (1, 4) and the integration interval has length 2, in the first nonzero interval of the convolution, which is [itex]1 \leq z \leq 3[/itex]
    this becomes
    [tex]h(z) = \frac{1}{6} \int_{1}^{z}du = \frac{z-1}{6}[/tex]

    Now suppose instead that Z = 2X + Y.

    Intuition tells me that there should be some change of variables to let me use convolution to derive a density function for this new Z, but if there is I can't find it. I have solved for the density function of this "new'' Z by finding its distribution function and then differentiating. This function has its first nonzero interval on [itex]1 \leq z \leq 4[/itex], and for this interval the differentiation method gives me
    [tex]h(z) = \frac{z-1}{12}[/tex].
    I'm pretty sure about this answer; I get the same result whichever direction I integrate.

    Is there a way to use a convolution to obtain this result?
    Last edited: Mar 16, 2007
  2. jcsd
  3. Mar 17, 2007 #2
    So far all I can see is this:
    I let
    Z = 2X + Y and (arbitrarily) T = X giving me
    X = T and Y = Z - 2T
    now the Jacobian of this transformation is
    J &= \left |\begin{array}{ll}
    \frac{\partial {x}}{\partial{z}} = 0 &\frac{\partial{x}} {\partial{t}} = 1 \\
    \frac{\partial{y}} {\partial{z}} = 1 &\frac{\partial{y}} {\partial{t}} = -2
    \end{array} \right | \\
    &= -1

    so |J| = 1.

    X and Y are independent so their joint density function [itex]j_1(x,y) = f(x) \cdot g(y)[/itex]
    and I should be able to write
    [tex]j_2(t,z) = j_1(x,y)|J| = j_1(t, z-2 t) = f(t) \cdot g(z-2t)[/tex]
    and then the marginal density function of z is
    m(z) &= \int_{-\infty}^{\infty} f(t) \cdot g(z-2t)dt \\
    &= \frac{1}{2} \int_0^2 g(z-2t)dt\\
    &= \frac{1}{2} \int_z^{z-4} g(u)\cdot (-\frac{1}{2}du)\\
    &= \frac{1}{4} \int_{z-4}^{z} g(u)du\\
    &= \frac{1}{12} \int_{z-4}^z du
    But since f is on the interval (0,2) while this interval of integration is length 4, the first non-zero interval of the result is
    [tex]h(z)= \frac{1}{12} \int_0^z du = \frac{z}{12} \mbox{ when } 0 < z < 2[/tex]

    but the correct result based on differentiating the distribution function for this interval was
    [tex]h(z) = \frac{z-1}{12} \mbox{ when } 1 < z < 4[/tex]
    so both the interval and the value of z are wrong.

    Can anyone spot where my error is?
  4. Mar 17, 2007 #3
    Convolution method only works for a sum. You'd need to find the distribution of 2X before you use it.
  5. Mar 17, 2007 #4
    Exactly what do you mean by that (the "distribution of 2X")? X is a continuous random variable. 2X is a function of X.

    The probability density of X is
    [tex]f(x) = \left\lbrace \begin{array}{ll}
    \frac{1}{2} &\mbox{ if } 0 < x < 2 \\
    0 &\mbox{otherwise}
    \end{array} \right. [/tex]

    Given that, is it valid to say that the distribution of 2X is
    [tex]f(2x) = \left\lbrace \begin{array}{ll}
    \frac{1}{2} &\mbox{ if } 0 < x < 1 \\
    0 &\mbox{otherwise}
    \end{array} \right. [/tex]

    or (considering "2x" as a name rather than 2 times x):
    [tex]f(2x) = \left\lbrace \begin{array}{ll}
    \frac{1}{2} &\mbox{ if } 1 < 2x < 2 \\
    0 &\mbox{otherwise}
    \end{array} \right. [/tex]

    I worked out the convolutions under both of these approaches (only for the first non-zero interval of each) and I got, for the former,
    [tex]h(z) = \frac{z}{12} \mbox{ when } 0 < z < 1[/tex]
    and for the latter,
    [tex]h(z) = \frac{z-2 }{6} \mbox{ when } 2< z < 3[/tex]

    so that doesn't work either, unless you had something else in mind.
    Last edited: Mar 17, 2007
  6. Mar 17, 2007 #5
    2x is a variable whose distribution depends on the distribution of x.


    You should verify this by either the distribution method or using jacobian transformations.
  7. Mar 17, 2007 #6
    Great! That's clear now:

    I'll let [itex]W = 2X[/itex]. Then [itex]X = \frac{1}{2}W \mbox{, and } \frac{dX}{dW} = \frac{1}{2}[/itex]
    [tex]f_W(w) = f_X(x)\frac{dX}{dW} = \frac{1}{2}f_x(x)[/tex]
    and when X = 0, W = 0 and when X = 2, W = 4 so
    [tex]f_W(w) = \frac{1}{2}f_X(x) = \left\lbrace \begin{array}{ll}
    \frac{1}{4} &\mbox{ if } 0 < w < 4 \\
    0 &\mbox{ otherwise} \end{array} \right.

    and the result of the convolution is now the exact value and interval that I expected. Thanks ZioX.
    Last edited: Mar 17, 2007
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