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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.

6 Answers 6

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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. $\int_0^2\mathbb{P}(Z>v)f_{X+Y}(v)\;dv$

  • 0
    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.2012-05-17
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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.2012-11-24
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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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    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.2012-05-18