# Probability that a new random variable is bigger than the i-th biggest random variable in a iid sequence?

Given a sequence of iid random variables $X_1, \dots, X_n, X_{n+1}$.

Given $i \in\mathbb N$ and $i \leq n$, how shall I find the probability of $X_{n+1}$ bigger than the $i$ th biggest one in $X_1, \dots, X_n$? I have no clue about how to tackle the problem.

Let's assume their common distribution is continuous, so that the probability of ties is zero.

Thanks!

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Let their pdf be $f(x)$ and cdf be $F(x)$
Then, pdf of $i^{th}$ order statistic $X_{(i)}$ is $h(x)=\frac{n!}{(i-1)!(n-i)!}(F(x))^{i-1}f(x)(1-F(x))^{n-i}$
Then required probability $=P(X_{n+1}\gt X_{(i)})=\int_{-\infty}^{\infty}P(X_{n+1}>x|X_{(i)}=x)h(x)dx$
Since, $X_{(i)}$ is a function of $X_1,X_2,...,X_n$ and $X_{n+1}$ is independent of those, therefore $X_{n+1}$ is independent of $X_{(i)}$ and you can replace $P(X_{n+1}>x|X_{(i)}=x)$ in integral with $P(X_{n+1}>x)=\int_x^{\infty}f(t)dt$