I have following exercise:
Let U, V, W be independent with uniform distribution on (0,1). We have indicators:
$$ I_1 = \left\{ \begin{array}{ll} 1 & U<V \\ 0 & \text{otherwise} \end{array} \right. $$
$$ I_2 = \left\{ \begin{array}{ll} 1 & V<W \\ 0 & \text{otherwise} \end{array} \right. $$
Calculate $var(I_1+I_2)$!
So far, I got those results: $$var(I_1+I_2) = var(I_1)+var(I_2)+2cov(I_1,I_2)$$ $$var(I_1) = E(I_1^2)-E(I_1)^2=P(U<V)P(U\geq V)$$ $$var(I_2) = E(I_2^2)-E(I_2)^2=P(V<W)P(V\geq W)$$ $$cov(I_1,I_2) = E(I_1 I_2)-E(I_1)E(I_2)=P(U<V, V<W)-P(U<V)P(V<W).$$
Now, I have problem calculating $P(U<V)$.
Can I say $P(U<V, V<W) = P(U<W)$ and $P(U<V) = P(V<W) = P(U<W)=p$. Then $var(I_1+I_2)=4p(1-p)$. But I still need to calculate $p$. How do I do that? Thank you.
It's true that $U<V, V<W \implies U<W,$ but $U<W \kern.6em\not\kern -.6em \implies U<V, V<W.$
For any of the probabilities of an inequality involving just two variables, for example $P(U<V),$ you might first consider the probability $P(U=V)$; and then in the case $U\neq V,$ consider which variable is more likely to be greater, $U$ or $V$? (This technique applies to discrete joint distributions too, not just continuous ones.)
There are various ways of computing $P(U<V,V<W).$ The most obvious is to consider the joint distribution of $(U,V,W)$ over the cube $\{(u,v,w) \mid 0< u< 1, 0< v< 1, 0< w< 1\}.$ This is a uniform distribution, that is, the density function of $(U,V,W)$ on this cube is $f(u,v,w) = 1.$ So the probability of an event $A$ is simply $$ P(A) = \int_A f(u,v,w) = \int_A 1.$$
For the event $A = \{U<V\} \cap \{V<W\}$ you can integrate $$ \int_A 1 = \int_{U<V,V<W} 1 = \int_?^?\int_?^?\int_?^? 1 \, du\,dv\,dw,$$ provided that you replace each "?" with the correct formula so that you only integrate over $(u,v,w)$ where $u < v < w.$
Alternatively, you might notice that the event $U<V, V<W$ tells you that the three variables $U,$ $V,$ and $W$ occur in that exact sequence along the number line, that is, $U<V<W.$ Now ask what is the probability that $W$ is the largest of the three variables; and given that $W$ is largest, which is more likely to be the next largest, $U$ or $V$? What about the possibility that there is a tie for the largest variable or the second largest variable? The answers to these questions let you compute $P(U<V, V<W).$
As another alternative, find the probability that there are any two variables amount $U,$ $V,$ and $W$ that are equal. Then list all the possible cases where there are no two equal variables. One such case is $U<V<W.$ What other cases are there, and which case is more likely than which other case? This way of computing $P(U<V, V<W)$ may have the simplest arithmetic, but it requires seeing certain symmetries of the problem correctly.