- #1

- 1,036

- 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?

[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: