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Let $X$, $Y$, and $Z$ be independently and identically distributed variables, each uniformly distributed between $0$ and $2$. What is the probability that $X+Y+Z\leq2$?

All the other answers on similar questions refer to various things like "convolutions" that I've never heard of or use integrals and change the limits from $[-\infty,+\infty]$ to other things etc. so I'm very confused and don't understand any of it. In particular, I thought we could just work out $P(X+Y\leq2-Z)=\frac{1}{8}(2-Z)^2$ and thus do $\int_{0}^{2}\frac{1}{8}(2-Z)^2 \text{d}Z$, and I don't understand why this doesn't give the right answer of $\frac{1}{6}$. Please would someone take the time to write out all the steps with full explanation for someone who has only basic knowledge of probability.

Prasiortle
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  • Consider $X/2,Y/2,Z/2$ which are i.i.d $U(0,1)$. – StubbornAtom Feb 15 '19 at 17:49
  • See a general result: https://math.stackexchange.com/questions/769545/volume-of-t-n-x-i-ge0x-1-cdotsx-n-le1?noredirect=1&lq=1 – StubbornAtom Feb 15 '19 at 18:04
  • You missed a factor of $\frac12$ in your integral. You need that factor to represent the density of $Z$. Think of it this way: if you wanted the probability that the sum is less than $6,$ you could write $P(X+Y<6-Z)=1,$ and then integrating over $Z$ from $0$ to $2$ would give the result $2$ when the correct total probability is $1.$ – David K Feb 15 '19 at 18:36

3 Answers3

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The sum of a given number of i.i.d. continuous uniform variables, on the support $[0,1]$ follows the Irwin-Hall distribution.

Since you are considering variables on the interval $[0,2]$, it is just a matter of re-scaling.

G Cab
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Idea:

  1. $X \sim {\rm Unif}([0,2])$
  2. In the required event, $0 \le Y \le 2 - X$.
  3. Determine the value of $Z$ similarily: $0 \le Z \le 2 - X - Y$.

$$\begin{aligned} & P(X+Y+Z \le 2) \\ &= \int_0^2 \int_0^{2-x} \int_0^{2-x-y} \left(\frac12\right)^3 \ dz \, dy \, dx \\ &= \left(\frac12\right)^3\int_0^2 \int_0^{2-x} (2-x-y) \, dy \, dx \\ &= \left(\frac12\right)^3\int_0^2 \int_0^{2-x} y \, dy \, dx \quad (\int_0^a f(x) dx = \int_0^a f(a-x) dx) \\ &= \left(\frac12\right)^3\int_0^2 \frac{(2-x)^2}{2} \, dx \\ &= \dots \text{(same trick)} \\ &= \left(\frac12\right)^3 \cdot \frac{2^3}{3!} = \frac16 \end{aligned}$$

Edit in response to OP's comment:

Consider $S = \{ (x,y,z) \in \Bbb{R}_+^3 \mid x + y + z \le 2\}$ and $S' = \{ (x,y,z) \in \Bbb{R}_+^3 \mid x \in [0,2], y \in [0,2-x], z \in [0,2-x-y]\}$. It suffices to show that $S = S'$. $S' \subseteq S$ is clear, and is left as exercise. To finish the argument, let $(x,y,z) \in S$. Then verify the criteria which define $S'$ componentwise.

  • $x \in [0,2]$ is clear
  • $y \in [0,2-x-z] \subseteq [0,2-x]$ is a bit tricky
  • $z \in [0,2-x-y]$ is also clear

Hence the upper and lower limits of the above iterated integral is justified.

GNUSupporter 8964民主女神 地下教會
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For those who prefer a geometric proof:

The random points are uniformly distributed in $X,Y,Z$ space within a cube of side 2.

$X+Y+Z=2$ defines a plane which intersects the cube at the points (2,0,0),(0,2,0) and (0,0,2). This is an equilateral triangle of side $2\sqrt 2$ and area $2\sqrt{3}$. The centre of the triangle is at (2/3,2/3,2/3) which is $2/\sqrt{3}$ from the origin.

The points for which $X+Y+Z\leq2$ are those in the triangular pyramid whose base is this triangle and whose apex is the origin. This has volume ${1 \over 3} A h={1 \over 3}{2\sqrt 3}{2 \over \sqrt 3}={4 \over 3}$.

Dividing by the total volume of 8 gives the probability ${1 \over 6}$

(The integrals are hidden in the formulae for the area ${ 1 \over 2} base \times height$ of a triangle and the volume ${1 \over 3} area \times height$ for a triangular pyramid.)

RogerJBarlow
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