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

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.

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$$\mathbb{P}(Z>X+Y)=\mathbb{E}[\mathbb{1}(Z>X+Y)]=\mathbb{E}[\mathbb{E}[\mathbb{1}(Z>X+Y)|X+Y]]=\mathbb{E}[\mathbb{P}(Z>X+Y|X+Y)],$$ 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. $\int_0^2\mathbb{P}(Z>v)f_{X+Y}(v)\;dv$

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I was puzzled by the first equality, but is the principle just that P(...) is some constant, and the expectation of a constant is a constant? If so that's a nice trick :) – maliky0_o May 17 '12 at 19:44
No, I don't think you understand it properly. The first equality involves indicator function $\mathbb{1}(Z>X+Y)$, which is $1$ on the event $Z>X+Y$ and $0$ on the event $Z\leq X+Y$. Then expectation of such indicator is just the probability of the event $Z>X+Y$, by definition of expectation. – Tom Artiom Fiodorov May 17 '12 at 19:48

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 $$P\{Z > X+Y\} = \frac{1}{3}\times (\text{area of base})\times(\text{altitude}) = \frac{1}{3}\times \frac{{3}}{2}\times1 = \frac{1}{6}.$$ 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 $$P\{Z > X+Y\}= \int_0^2{P}(Z>v)f_{X+Y}(v)\;dv = \int_0^1(1-v)\cdot v\;dv = \left.\frac{v^2}{2}-\frac{v^3}{3}\right|_0^1 = \frac{1}{6}.$$

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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.

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This tries to address the question asked, which is to explain the identity written in the post, and not to compute $P(Z\gt X+Y)$ in the specific situation described in the post. – Did Nov 24 '12 at 11:28

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

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I think your question best understood using two discrete random variables. Suppose you have two random variables $X$ and $Y$ taking values $0,1,2,\ldots,\infty$. Now you are asked to compute the probability of the event $A = X > Y$.

So, $$\begin{eqnarray} P(A) &=& P(X>Y)\\ \end{eqnarray}$$ Here both $X$ and $Y$ are random. To compute this probability we need the notion of conditional probability. Here it is: $$P(A \cup B) = P(A|B) \times P(B)$$

Now, come to the original problem. We first fix the value of any one random variable, say, $Y = y$. Clearly, $y$ is any value from $0,1,2,\ldots,\infty$, but $y$ can't take these values simultaneously. Now we compute $P(X > y|Y = y)$ and $P(Y = y)$.

Hence $P(X > Y)$ is nothing but $P(X > y|Y = y) \times P(Y = y)$. But we have probabilities of so many events like this for each and every possible value of $y$, again each of these events are mutually exclusive, because occurrence of any one, say, $y = 1$ prevents the occurrence of others i.e. $y = i, i \neq 1$. Therefore, to get the required probability, we need to sum up the probabilities for each of the m.e. events. Thus finally we get, $$P(X > Y) = \sum_{y = 0}^{\infty} \left[P(X > y| Y = y) \times P(Y = y)\right]$$

If you are familiar with the basic definition of expectation of random variable, then previous expression is actually, $$\begin{eqnarray} P(X > Y) &=& \sum_{y = 0}^{\infty} \left[value \times \text{corresponding probability}\right]\\ P(X > Y) &=& E\left[P(X > y| Y = y)\right] \end{eqnarray}$$

Now, to make this result suitable for continuous variable, just replace the sum by integration w.r.t $y; (0 \leq y < \infty)$ and $P(Y = y)$ by $f_Y(y)$ i.e. density function of $Y$ at the point $y$.

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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.

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Michael, could you elaborate on how this differs from what Artiom Fiodorov wrote, namely $\int_0^2\mathbb{P}(Z>v)f_{X+Y}(v)\;dv$ – Dilip Sarwate May 17 '12 at 23:20
Sure. He got it by integrating over the entire 3D space that he point out is a tetrahedron where the density is nonzero. In my case I got to the last step explictly using the expectation of the conditional probability which shows how that rule gets used in the derivation. He went straight to the last step without using this or at least without explaining that he used this. – Michael Chernick May 18 '12 at 4:54
The he who integrated over the entire 3D space and used tetrahedrons is me; the Artiom Fiodorov who explained that $$P(Z>X+Y)=E[\mathbb{1}(Z>X+Y)]=E[E[\mathbb{1}(Z>X+Y)|X+Y]]=E[P(Z>X+Y|X+Y)],$$ and then showed that this works out to $\int_0^2P(Z>v)f_{X+Y}(v)\;dv$ in his answer is different. The request was that you elaborate on how your answer differs from Artim's answer. – Dilip Sarwate May 18 '12 at 13:09
@DilipSarwate After reading a little more carefully I see that you did bothe the three dimensional solution and finished up Artim's. He wrote out the the solution in terms of an integral without deriving the density for X+Y. Taking your completion of it with what he does is basically my solution. I gave a little more of a detailed explanation that showed why the integral from 1 to 2 is zero. There is some redundancy here and it would be better to cleans this up just showing the two solutions completely. – Michael Chernick May 18 '12 at 22:00
That could go into two answers. One being the first part of yours and the other combining mine with his. I am not sure what is the best way to edit the answers to tighten things up. – Michael Chernick May 18 '12 at 22:00