Given two independent variables with these simple density functions:(adsbygoogle = window.adsbygoogle || []).push({});

[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?

**Physics Forums - The Fusion of Science and Community**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Convolution math problem

Loading...

Similar Threads - Convolution math problem | Date |
---|---|

I Question about simplifying Sigma notation | Feb 11, 2018 |

I Deriving convolution | Sep 24, 2016 |

A Cauchy convolution with other distribution | Jul 18, 2016 |

A Convolution questions | Jul 6, 2016 |

Convolution of Time Distributions | Feb 17, 2016 |

**Physics Forums - The Fusion of Science and Community**