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Let $U_1$, $U_2$ and $U_3$ be i.i.d. $\mathsf {Uniform}[0,1]$ random variables. Find $P(U_1+U_2 > U_3)$

Here the answer states that the probability is $\frac{5}{6}$. However I am quite confused as to how we obtain such an awnser. A hint is to use conditional expectation.

Any advice or explanation is apreciated.

I saw the following link :Probability that $\max(U_1,U_2) > U_3$ for independent uniform random variables $U_i$

And the key argument was the symetry of the system.

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rannoudanames
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  • Check this http://math.stackexchange.com/questions/610562/if-u1-u2-u3-are-iid-uniform0-1-then-what-is-the-probability-of-u1u2u3 – vkonton Dec 14 '16 at 22:08

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The PDF of $U_1+U_2$ is supported on $[0,2]$ and equals $x$ on $[0,1]$ and $2-x$ on $[1,2]$.
It follows that: $$\begin{eqnarray*}\mathbb{P}[U_1+U_2>U_3] &=& \int_{0}^{1}\int_{u}^{2}(1-|v-1|)\,dv\,du\\&=&\int_{0}^{1}\left(\frac{1}{2}+\int_{u}^{1}v\,dv\right)\,du \\&=&\int_{0}^{1}\frac{2-u^2}{2}\,du\\&=&1-\frac{1}{6}=\color{blue}{\frac{5}{6}}.\end{eqnarray*}$$

Jack D'Aurizio
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  • I see how what you write is true, however the method is not what i looking for. I am however getting close to something, will post if i find – rannoudanames Dec 14 '16 at 23:22
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    @rannoudanames $P(U_1+U_2<U_3) = E[1_{{U_1+U_2>U_3}}]$, then use the law of total expectation to get $E[1_{{U_1+U_2>U_3}}] = E[E[1_{{U_1+U_2>u}}\mid U_3 = u]]$. Perhaps these intermediate steps help you see how the first integral is derived from conditional expectation? – Therkel Dec 15 '16 at 06:49
  • That makes sense! – rannoudanames Dec 15 '16 at 17:17