TITLE: How to solve $P(\max\{X_1,X_2,...,X_n\}>b|X_1=a)$ when $X_i$ is uniformly distributed?
QUESTION [0 upvotes]: Original problem:
Let $X_1,X_2,...,X_n$ be i.i.d. random variables, each uniformly distributed over [0, 1]. Let $V=\max\{X_1,X_2,...,X_n\}$. Determine $P(V>b|X_1=a)$, when $a,b\in [0, 1]$.
My questions:

How to evaluate it?
$P(V>b|X_1=a)=1-P(X_2\le b)P(X_3\le b)...P(X_n\le b)P(X_1\le b|X_1=a)$, so when $a<b$, $P(X_1\le b, X_1=a)=P(X_1\le a)$, but it should be $P(X_1\le b,X_1=a) \le P(X=a) = 0$, so $P(X_1\le b,X_1=a)=0$... Where is wrong?

REPLY [1 votes]: If $a>b$ then we already know that the probability of interest is $1$. Otherwise for $a\leq b$,
$$
\begin{align*}
P(V>b\mid X_1=a)&=1-P(V\leq b\mid X_1=a) \\
&=1-P(X_1\leq b,\dots,X_n\leq b\mid X_1=a) \\
&=1-\prod_{i=1}^nP(X_i\leq b\mid X_1=a) \\
&=1-\prod_{i=2}^nP(X_i\leq b\mid X_1=a) \\
&=1-\prod_{i=2}^nP(X_i\leq b) \\
&=1-b^{n-1}.
\end{align*}
$$
{"set_name": "stack_exchange", "score": 0, "question_id": 2144943}