Find $ P(Z>X+Y)$ where $X,Y,Z \sim U(0,1)$ independently

Question

I'm trying to follow a line in a derivation for $P(Z>X+Y)$ where $X,Y,Z$ are independent continuous random variables distributed uniformly on $(0,1)$.

I've already derived the pdf of $X+Y$ using the convolution theorem, but there's a line in the answer that says:

$P(Z>X+Y) = \mathbb{E}[\ P(Z>X+Y\ |\ X+Y )\ ]$ where $\mathbb{E}$ is the expectation.

I'm not familiar with this result. Could anyone give a pointer to a similar result if one exists?

Thanks.

Answer 1

$$\mathbb{P}(Z>X+Y)=\mathbb{E}[\mathbb{1}(Z>X+Y)]=\mathbb{E}[\mathbb{E}[\mathbb{1}(Z>X+Y)|X+Y]]=\mathbb{E}[\mathbb{P}(Z>X+Y|X+Y)],$$ where second equality is the following property of conditional expectation: $$\mathbb{E}[\mathbb{E}[X|Y]]=\mathbb{E}[X]$$ Intuitively, now that you know distribution of $X+Y$, you just need to "range"$^1$ through the values of $X+Y$, and find the probability of $Z>X+Y$ for each such value. This is exactly the expectation of the probability.

$^1$integrate against the density, i.e. $\int_0^2\mathbb{P}(Z>v)f_{X+Y}(v)\;dv$

Answer 2

A partial justification can be found in the Wikipedia entry on the Law of Total Probability.

Answer 3

This is not an answer to your question about the justification for the equation that is puzzling you, but I think the geometrical method described below for solving the problem that may give you a different insight into the calculation of the desired probability $P\{Z > X+Y\}$.

The random point $(X,Y,Z)$ is uniformly distributed in the interior of the unit cube with diagonally opposite vertices $(0,0,0)$ and $(1,1,1)$. The cube has unit volume and so the probability that $(X,Y,Z)$ is in some region is just the volume of that region. Thus, $P\{Z > X+Y\}$ is the volume of the tetrahedron with vertices $(0,0,0)$, $(1,0,1)$, $(0,1,1)$ and $(0,0,1)$. If we think of this as an inverted pyramid whose base is the right triangle with vertices $(1,0,1)$, $(0,1,1)$ and $(0,0,1)$ and apex $(0,0,0)$ is at altitude $1$ "above" the base, then since the base has area $\frac{1}{2}$, we get the volume as $$P\{Z > X+Y\} = \frac{1}{3}\times (\text{area of base})\times(\text{altitude}) = \frac{1}{3}\times \frac{{1}}{2}\times1 = \frac{1}{6}.$$ Of course, if you have already computed the density of $X+Y$, then it is straightforward to use the result given by Artiom Fiodorov to get $$P\{Z > X+Y\}= \int_0^2{P}(Z>v)f_{X+Y}(v)\;dv = \int_0^1(1-v)\cdot v\;dv = \left.\frac{v^2}{2}-\frac{v^3}{3}\right|_0^1 = \frac{1}{6}.$$

Answer 4

I don't know if this helps since Dilip has given the answer, but the distribution of X+Y is triangular on [0,2] (isosceles with peak at X+Y = 1). So P(Z>X+Y) is the probability that a uniform on [0,1] is larger than the triangular random variable on [0,2]. If X+Y>1 then Z cannot be >X+Y and the probability that X+Y is greater than 1 is 1/2. Now this is where taking the expectation fo the conditional probability helps in my proof. P{Z>X+Y) =E[P(Z>X+Y|X+Y)]= ∫u P(Z>u|X+Y=u)du =∫u P(Z>u)du where u is integrated from 0 to 1. The condition X+Y=u gets dropped because Z is independent of X+Y. P(Z>u)=1-u for 0<=u<=1. hence P(Z>X+Y) =∫u(1-u)du = 1/6. Just as Dilip showed.

Answer 5

Replacing the question in a larger context might help. Here is a result:

For every event $A$ in $(\Omega,\mathcal F,\mathbb P)$ and every sigma-algebra $\mathcal G\subseteq\mathcal F$, $\mathbb P(A)=\mathbb E(\mathbb P(A\mid \mathcal G))$.

To see this, recall that $U=\mathbb P(A\mid \mathcal G)$ is the unique (up to null events) random variable such that $\mathbb E(U;B)=\mathbb P(A\cap B)$ for every $B$ in $\mathcal G$. In particular, $B=\Omega$ yields $\mathbb E(U)=\mathbb P(A)$, as claimed above.

In your setting, $A=[Z\gt X+Y]$ and $\mathcal G$ is the sigma-algera generated by the random variable $X+Y$ hence $\mathbb P(\ \mid \mathcal G)=\mathbb P(\ \mid X+Y)$ by definition.