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Suppose $X$ and $Y$ are iid's uniform distributions on [0,1]. Then $Z=X+Y$.

Reading through other similar questions such as density of sum of two uniform random variables $[0,1]$, I'm still struggling to understand how exactly the interval $0<z<2$ is reached.

It seems to me just to start this problem it is necessary to define the interval of $z$. I assume that if $X$ and $Y$ can only take on values from $[0,1]$, then at most $z$ can be $2$. Is this the correct way to think about this? More specifically I'm referring to the need of two cases, $0<z<1$ and $1<z<2$, and how they are obtained.

pajkatt
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1 Answers1

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The variables can each be between $0$ and $1$, so the smallest the sum can be is $0$ and the largest the sum can be is $2$.

As for the actual distribution, the reason for why the distribution of the sum of two independent random variables is the convolution of the distributions is because: $$ P(Z = z) = \sum_x P(X = x) \cdot P(Y = z - x) $$ where the sum is taken over all possible values of $x$. When it comes to continuous random variable, we must use the probability density function, and so: $$f_Z(z)= \int_{-\infty}^\infty f_X(x) f_Y(z-x) \, dx $$ This integral is the convolution of $f_X$ and $f_Y$.

In the case of these two distributions in this specific problem, the convolution looks like this: https://en.wikipedia.org/wiki/Convolution#/media/File:Convolution_of_box_signal_with_itself2.gif

This section explaining the visual interpretation may help as well: https://en.wikipedia.org/wiki/Convolution#Visual_explanation

Manuel Guillen
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