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Let $X$ and $Y$ be $iid$ uniformly distributed random variables over the interval $[0,1]$. We know by convolution that the distribution of $Z=X+Y$ is given by:

$$f(z) = \left \{ \begin{array}{ccc} z & \text{if} & 0\leq z \leq 1 \\ 2-z & \text{if} & 1 \leq z \leq 2 \end{array} \right. $$

What would happen if the transformation were a linear combination? $Z=c_1 X+ c_2 Y$. Can I still use convolution to find this transformation?

$$f(z) = \int_{-\infty}^{+\infty}f_X\left (\frac{z - c_2 y}{c_1} \right)f_Y(y)dy$$ But $f_Y(y) = 1$ since it is uniform, $$f(z) = \int_0^1f_X\left (\frac{z - c_2 y}{c_1} \right)dy$$

I am stuck. Are the bounds correct? What should I do after? Thanks

cgo
  • 1,810

2 Answers2

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w.l.o.g. (justified by $X$ and $Y$ being identically distributed) there are three distinct cases

i) $c_1, c_2 > 0$;

ii) $c_1>0 , c_2 < 0$;

iii) $c_1 , c_2 < 0$.

I will solve case (i), leaving the other two for you to develop analogously. As you suspect, we can solve each of these cases using convolution. For case (i) assume, again w.l.o.g., that $c_1 > c_2$. Note that $c_1X$ is uniformly distributed, independently of $c_2Y$, on the interval $[0,c_1]$, and analogously for $c_2Y$. This means that the respective pdfs are $$ f_{c_1X}(x){}={}\dfrac{1}{c_1}{\bf{1}}_{\left\{0<x<c_1\right\}}\,\,\,\,\mbox{ and }\,\,\,\,f_{c_1Y}(y){}={}\dfrac{1}{c_2}{\bf{1}}_{\left\{0<y<c_2\right\}}\,. $$

Using this, observe that the non-zero range of the random variable $Z=c_1X+c_2Y$ is the interval

$$ 0<Z<c_1+c_2\,, $$

within which we have 3 disjoint sub-intervals (and, therefore, three convolutions to solve). You can show that:

(i) for $0\le z \le c_2$, $$ \int\limits{\bf{1}}_{\left\{0<y<z\right\}}\,f_{c_1X}\left(z-y\right)f_{c_2Y}\left(y\right)\,\mathrm{d}y{}={}\int\limits_{0}^{z}\,f_{c_1X}\left(z-y\right)f_{c_2Y}\left(y\right)\,\mathrm{d}y{}={}\dfrac{z}{c_1c_2}\,; $$

(ii) for $c_2 < z \le c_1$,

$$ \int\limits{\bf{1}}_{\left\{z-c_2<x<z\right\}}\,f_{c_2Y}\left(z-x\right)f_{c_1X}\left(x\right)\,\mathrm{d}x{}={}\int\limits_{z-c_2}^{z}\,f_{c_2Y}\left(z-x\right)f_{c_1X}\left(x\right)\,\mathrm{d}x{}={}\dfrac{c_2}{c_1c_2}\,; $$

(iii) for $c_1 < z \le c_1+c_2$,

$$ \int\limits{\bf{1}}_{\left\{z-c_2<x<c_1\right\}}\,f_{c_2Y}\left(z-x\right)f_{c_1X}\left(x\right)\,\mathrm{d}x{}={}\int\limits_{z-c_2}^{c_1}\,f_{c_2Y}\left(z-x\right)f_{c_1X}\left(x\right)\,\mathrm{d}x{}={}\dfrac{c_1-z+c_2}{c_1c_2}\,. $$

This shows that

$$ f_{Z}(z){}={\bf{1}}_{\left\{0\le z \le c_2\right\}}\dfrac{z}{c_1c_2} {}+{}{\bf{1}}_{\left\{c_2 < z \le c_1\right\}}\dfrac{c_2}{c_1c_2} {}+{}{\bf{1}}_{\left\{c_1 < z \le c_1+c_2\right\}}\dfrac{c_1-z+c_2}{c_1c_2}\,. $$

ki3i
  • 5,092
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The formulas for $f(z)$ appear to be correct. But in the general case $c_1X + c_2Y$, the result will not be two piecewise linear segments like the case for $X + Y$. In general (unless $c_1 = c_2$) you will have three segments.

Try to evaluate $f(z)$ at a few different values of $z$ to see what the shape of that density is. The tricky part is to determine, for each value of $z$, which subrange of $y$ within $[0,1]$ will make $f_X\left(\frac{z-c_2y}{c_1}\right) = 1$. You might want to consider the case $c_1 < c_2$ separately from the case $c_1 > c_2.$

David K
  • 98,388
  • Is there a name for the distribution that arises from a linear combination of uniform distributions? The Irwin-Hall distribution only covers $c_i = 1$. – a06e Nov 08 '16 at 16:38
  • I don't know a name. Irwin-Hall has a name because it gets used a lot. The case $c_i \neq c_j$ is more complicated, so unless it has some important applications for which we can't use Irwin-Hall, I imagine it just wouldn't be mentioned often and it would be useless to name it. But maybe there is an frequently-used application that I don't know, and the people who do that application have a name for the distribution. – David K Nov 08 '16 at 16:48