Probability conditional on inequality

Question

Let $X_1,X_2,....,X_{n+1}$ independent uniform variables on the unit interval $[0,1]$, $V:=\max(X_1,X_2,...,X_n)$ and $c \in [0,1].$

What is $\textrm{Prob}(X_{n+1}<V|V<c)$?

Initially I thought this would be equivalent to $$\textrm{Prob}(X_{n+1}<V<c)=\int_0^cnz^ndz=\frac{n}{n+1}z^{n+1},$$

but I am concerned that equivalence may not hold.

Answer 1

Since $V < c$ means $X_1, \ldots, X_n < c$, the probability of $X_i$ conditioned on $V < c$ is uniform $$ \Pr(X_i \mid V<c) = 1 / c, $$ and $$ X_i \mid V < c \sim U(0,c), $$ for $i = 1\ldots n$.

Consider $V := \max(X_1, \ldots, X_n) = X_j$, for $1 \leq j \leq n$, which follows that $$ \begin{align} &\Pr(V < v \mid V < c) \\ =& \prod_j \Pr(X_j < v \mid V <c) \\ =& (v / c)^n. \end{align} $$ and $$ \Pr(V=v \mid V<c) = \frac{d}{dv}Pr(V < v \mid V < c) =nv^{n-1}/c^n. $$

Given that $X_1, \ldots, X_n, X_{n+1}$ are independent, it follows that $$ \Pr(X_{n+1} < V \mid V <c) = \Pr(A < B), $$ where $j = 1\ldots n$ , $A \sim U(0,1)$, and $B = V\mid V<c$.

From Probability and uniform distribution, we know that $$ \begin{align} &\Pr(B>A) \\ =& \int_{0}^1 P(B>A | A=x) f_{A}(x) dx \\ =& \int_{0}^c P(B>A | A=x) f_{A}(x) dx \qquad \text{($A$ must be lesser than $B$)}\\ =& \int_{0}^c (1-(x / c)^n) \cdot 1 dx \\ =& \left. c - \frac{x^{n+1}}{(n+1)c^n} \right\vert_0^c \\ =& \frac{n}{n+1}c \end{align} $$ Therefore, $$ \Pr(X_{n+1} < V \mid V <c) = \frac{n}{n+1}c. \tag{1} $$

Moreover, eq. $(1)$ can be validated by the conditional probability formula $$ \begin{align} &\Pr(X_{n+1} < V \mid V < c) \Pr(V < c) \\ =&\frac{n}{n+1}c \cdot c^n \\ =&\Pr(X_{n+1} < V < c). \end{align} $$

Answer 2

The formal meaning of $P(X_{n+1}<V|V<c)$ is $E(1_{X_{n+1}<V}|1_{V<c})$. The conditional expectation writes as $f(1_{V<c}) = f(1)1_{V<c} + f(0)1_{V\geq c}$ for some measurable $f$.

The $\sigma$-algebra generated by $1_{V<c}$ is $\{\emptyset, (V<c), (V\geq c), \Omega \}$ thus the defining equations of the conditional expectation write as $$E(1_{V<c} 1_{X_{n+1}<V}) = E( 1_{V<c} (f(1)1_{V<c} + f(0)1_{V\geq c})) = f(1)E( 1_{V<c} ) = f(1)P(V<c)$$

$$E(1_{V\geq c} 1_{X_{n+1}<V}) = E( 1_{V\geq c} (f(1)1_{V<c} + f(0)1_{V\geq c})) = f(0)E( 1_{V\geq c} ) = f(0)P(V\geq c)$$

Thus $\displaystyle f(1)= \frac{P(X_{n+1}<V<c)}{P(V<c)}$ and $\displaystyle f(0)= \frac{P((X_{n+1}<V) \cap (V\geq c))}{P(V\geq c)}$.

Both quantities are easily computable by integral means.