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 $E$ := {$U_1 \geq U,U_2 \geq U,U_3 &lt; U,U_4 \geq U, U_5 &lt; U,U_6 \geq U,U_7 \geq U$}.

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!
Recall that

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

Observe that


$$\mathbb{P}[\max\{U_3, U_5\} \le u] = (\max\{\min\{u,1\},0\})^2$$
$$u = \max\{\min\{u,1\},0\}, u \in [0,1]$$
$$\mathbb{P}[\min\{U_1, U_2, U_4, U_6, U_7\} \le u] = 1 - (1 - \mathbb{P}[U_1 \le u])^5$$
$$E = \{\max\{U_3, U_5\} &lt; U \le \min\{U_1, U_2, U_4, U_6, U_7\} \}$$


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

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


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


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

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

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

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

$$= 1/3$$

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


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


Integrate that if you want. :P


So how does 4 relate to 5 and 6 in terms of computing $P(E)$?
Recall that

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

Observe that


$$\mathbb{P}[\max\{U_3, U_5\} \le u] = (\max\{\min\{u,1\},0\})^2$$
$$u = \max\{\min\{u,1\},0\}, u \in [0,1]$$
$$\mathbb{P}[\min\{U_1, U_2, U_4, U_6, U_7\} \le u] = 1 - (1 - \mathbb{P}[U_1 \le u])^5$$
$$E = \{\max\{U_3, U_5\} &lt; U \le \min\{U_1, U_2, U_4, U_6, U_7\} \}$$


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

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


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


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

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

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

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

$$= 1/3$$

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


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


Integrate that if you want. :P


So how does 4 relate to 5 and 6 in terms of computing $P(E)$?