Let $X$ ~ Uniform$[a,b]$ and $Y$ ~ Uniform$[c,d],$ where $a\le b\le c\le d.$ Find the probability density of $Z = X + Y.$

I know I have to use the convolution formula

$$f_Z(z) = \int_{-\infty}^\infty f_X(z-y)f_Y(y)\,\mathrm dy.$$

However, I am struggling to find the region I must integrate over. I know for $z \le a+c$ and $z \gt b+d$ that $f_Z(z) = 0.$ But for the middle terms I do not know what to do. Any help is appreciated.

  • $\begingroup$ It may be worth also breaking the integration at $z=a+d$ and at $z=b+c$ (which might be in either order) $\endgroup$
    – Henry
    Jul 24, 2023 at 7:43
  • $\begingroup$ I'm sorry, but could you explain what you might mean by "middle terms"? The region of integration is always the entire set of real numbers; there are no exceptions. In this case you perform the integration by recognizing that $f_X$ is zero outside the interval $[a,b]$ and $f_Y$ is zero outside $[c,d].$ $\endgroup$
    – whuber
    Jul 24, 2023 at 13:13
  • $\begingroup$ BTW, if you would draw a picture of the support of $(X,Y)$ and some contour lines of the function $X+Y,$ you can (literally) read the answer off that diagram. No calculation (not even any algebra) is needed at all. $\endgroup$
    – whuber
    Jul 24, 2023 at 14:06

1 Answer 1


Plot the contours (level curves) of the function $(x,y)\to x+y$ over the support of this distribution.

Here, to illustrate, is such a diagram with $(a,b)=(0,2)$ and $(c,d)=(3,4):$

enter image description here

The value of $z=x+y$ ranges from $a+c$ to $b+d$ as you sweep from the lower left to upper right (the direction given by the vector $(1,1),$ since $x+y = (x,y)\cdot (1,1)^\prime$). As it does so, since the distribution of $(x,y)$ is uniform, the lengths of the contours are proportional to the probability density of $z.$

The diagram makes it clear that those lengths:

  • (Region I) Increase linearly from $0$ as $z$ goes from $a+c$ (in the lower left corner) to $a+d$ (in the upper left corner),

  • (Region II) then remain constant as $z$ goes from $a+d$ to $b+c,$

  • (Region III) then decrease linearly to $0$ until $z$ reaches $b+d$ (in the upper right corner).

This description would change slightly were the rectangle vertically oriented (the order of $a+d$ and $b+c$ gets reversed), but the pattern remains the same. Consequently, $z$ has a trapezoidal distribution and we can plot its density function without further ado:

enter image description here

We don't need to label the vertical axis because the total area must be $1.$ In this diagram that implies the height of the trapezoid is $1/(b-a).$ You can easily write down equations for the five parts of this graph (over the intervals delimited by $-\infty, a+b, a+d, b+c, b+d,$ and $+\infty$) from this information.

This graphical solution will guide you in articulating a formal answer in terms of integrals.


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