# Convolution math problem

1. Mar 16, 2007

### gnome

Given two independent variables with these simple density functions:

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

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

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

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

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 $1 \leq z \leq 4$, and for this interval the differentiation method gives me
$$h(z) = \frac{z-1}{12}$$.
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. Mar 17, 2007

### gnome

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
$$\begin{array}{ll} J &= \left |\begin{array}{ll} \frac{\partial {x}}{\partial{z}} = 0 &\frac{\partial{x}} {\partial{t}} = 1 \\ \frac{\partial{y}} {\partial{z}} = 1 &\frac{\partial{y}} {\partial{t}} = -2 \end{array} \right | \\ \\ &= -1 \end{array}$$

so |J| = 1.

X and Y are independent so their joint density function $j_1(x,y) = f(x) \cdot g(y)$
and I should be able to write
$$j_2(t,z) = j_1(x,y)|J| = j_1(t, z-2 t) = f(t) \cdot g(z-2t)$$
and then the marginal density function of z is
\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*}
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
$$h(z)= \frac{1}{12} \int_0^z du = \frac{z}{12} \mbox{ when } 0 < z < 2$$

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

Can anyone spot where my error is?

3. Mar 17, 2007

### ZioX

Convolution method only works for a sum. You'd need to find the distribution of 2X before you use it.

4. Mar 17, 2007

### gnome

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

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

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

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,
$$h(z) = \frac{z}{12} \mbox{ when } 0 < z < 1$$
and for the latter,
$$h(z) = \frac{z-2 }{6} \mbox{ when } 2< z < 3$$

so that doesn't work either, unless you had something else in mind.

Last edited: Mar 17, 2007
5. Mar 17, 2007

### ZioX

2x is a variable whose distribution depends on the distribution of x.

$$f_{(2x)}(2x)=\frac{1}{4}I_{(0,4)}$$

You should verify this by either the distribution method or using jacobian transformations.

6. Mar 17, 2007

### gnome

Great! That's clear now:

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

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

Last edited: Mar 17, 2007