# Probability that $U_1 \geq U$, $U_2 \geq U$, $U_3 < U$, $U_4 \geq U$, $U_5 < U$, $U_6 \geq U$, $U_7 \geq U$, for i.i.d. uniform $U_k$s and $U$

Let $U,U_1,U_2,...$ be independant, on [0,1] uniform distributed random variables.

Let $E$ := {$U_1 \geq U,U_2 \geq U,U_3 < U,U_4 \geq U, U_5 < U,U_6 \geq U,U_7 \geq U$}.

Find the probabiliy $P(E)$ and then find the probability $P(U_8 \geq U |E)$.

I am struggling with this task and therefore need help, thanks!

• Where have you got with the first part, finding p(E), which is much easier than the 2nd part? What is $p(U_i<U)$? – almagest Jun 20 '16 at 9:27
• i couldnt solve it, this kind of task is totally new to me , calculating probabilites without numbers – jordan178 Jun 20 '16 at 9:30
• There are two ways of finding $p(U_i<U)$. One is symmetry, the other is to integrate. – almagest Jun 20 '16 at 9:32
• do you mind posting a solution, maybe I will then be able to follow – jordan178 Jun 20 '16 at 9:40
• does min{$U_3,U_5$} $< U \leq$ min{$U_1,U_2,U_4,U_6,U_7$} help me? – jordan178 Jun 20 '16 at 9:51

Recall that

$$\mathbb{P}[U \le u] = \max\{\min\{u,1\},0\})$$

Observe that

1. $$\mathbb{P}[\max\{U_3, U_5\} \le u] = (\max\{\min\{u,1\},0\})^2$$

2. $$u = \max\{\min\{u,1\},0\}, u \in [0,1]$$

3. $$\mathbb{P}[\min\{U_1, U_2, U_4, U_6, U_7\} \le u] = 1 - (1 - \mathbb{P}[U_1 \le u])^5$$

4. $$E = \{\max\{U_3, U_5\} < U \le \min\{U_1, U_2, U_4, U_6, U_7\} \}$$

$$= \{\max\{U_3, U_5\} \color{red}{\le} U \le \min\{U_1, U_2, U_4, U_6, U_7\} \}$$

$$= \{\max\{U_3, U_5\} \le U \} \cap \{U \le \min\{U_1, U_2, U_4, U_6, U_7\} \}$$

1. $$P(\{\max\{U_3, U_5\} \le U \}) = P(\{\max\{U_3, U_5\} - U \le 0\})$$

$$= \int_0^1 \int_0^u f_{\max\{U_3, U_5\},U} du_{3,5}du$$

$$= \int_0^1 \int_0^u f_{\max\{U_3, U_5\}}f_{U} du_{3,5}du$$

$$= \int_0^1 \int_0^u f_{\max\{U_3, U_5\}}(1) du_{3,5}du$$

$$= \int_0^1 \int_0^u (2u_{3,5})(1) du_{3,5}du$$

$$= 1/3$$

Or simply note that

$$P(\{\max\{U_3, U_5\} - U \le 0\}) = P(\{\max\{U_3, U_5, U\} = U\})$$

where the RHS is 1/3 by symmetry

1. $$P(\{U \le \min\{U_1, U_2, U_4, U_6, U_7\} \}) = P(\{U = \min\{U, U_1, U_2, U_4, U_6, U_7\} \}) = 1/6$$

Integrate that if you want. :P

1. So how does 4 relate to 5 and 6 in terms of computing $P(E)$?
• "So how does 4 relate to 5 and 6 in terms of computing P(E)?" No. Idea. Whatsoever. (This whole post seems pure spam.) – Did Jul 2 '16 at 19:24