Find $ P(Z>X+Y)$ where $X,Y,Z \sim U(0,1)$ independently 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.
 A: 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
$$P\{Z > X+Y\}= \int_0^2{P}(Z>v)f_{X+Y}(v)\;dv
= \int_0^1(1-v)\cdot v\;dv = 
\left.\frac{v^2}{2}-\frac{v^3}{3}\right|_0^1 = \frac{1}{6}.$$
A: $$\mathbb{P}(Z>X+Y)=\mathbb{E}[\mathbb{1}(Z>X+Y)]=\mathbb{E}[\mathbb{E}[\mathbb{1}(Z>X+Y)|X+Y]]=\mathbb{E}[\mathbb{P}(Z>X+Y|X+Y)],$$
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. $\int_0^2\mathbb{P}(Z>v)f_{X+Y}(v)\;dv$
A: 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.
A: A partial justification can be found in the Wikipedia entry on the Law of Total Probability.
A: 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 $y$.
A: 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.
