Convolution math problem

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  • #1
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Main Question or Discussion Point

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:

Answers and Replies

  • #2
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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
[tex]\begin{array}{ll}
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
\end{array}[/tex]

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
[tex]\begin{align*}
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
\end{align*}
[/tex]
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?
 
  • #3
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Convolution method only works for a sum. You'd need to find the distribution of 2X before you use it.
 
  • #4
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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.
 
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  • #5
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2x is a variable whose distribution depends on the distribution of x.

[tex]f_{(2x)}(2x)=\frac{1}{4}I_{(0,4)}[/tex]

You should verify this by either the distribution method or using jacobian transformations.
 
  • #6
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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]
so
[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.
[/tex]


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

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