Probability distribution of $X_N$, $N = min\{n \geq 2: X_n = $ second largest of $ X_1, \ldots, X_n\}$ Given a sequence $X_1, X_2, \ldots$ of independent, continuous random variables with the same distribution function $F$ and density function $f$, let $N = min\{n \geq 2: X_n = $ second largest of $ X_1, \ldots, X_n\}$.
Denote $X_N$ as the first random variable which, at the time of observation, is the second largest of those observed so far.  Find the probability density function $f_{X_N}(x)$.
I'm extremely lost on this problem; all I've come up with thus far is to look at random variables $X_M$, the maximum of all observed so far, and $X_{M-1}$, the 'second-to-maximum' but I have been unable to find their respective distribution/density functions.
Thanks in advance for any and all help.
 A: Let $f(x)$ be the density function of the $X_i$, and $F(x)$ their cdf.
Let $Z$ be the maximum of the $X_i$. For completeness we find the distribution of $Z$, though that is likely familiar to you.
The event $Z\le z$ happens precisely if all the $X_i$ are $\le z$. This has probability $(F(z))^n$. Thus
$$F_Z(z)=(F(z))^n.$$
For the density function of $Z$, differentiate. We get $f_Z(z)=nf(z)(F(z))^{n-1}$.

Let $Y$ be the second largest of the $X_i$.  We give a highly informal derivation of the density function $f_Y(y)$ of $Y$.
Let $dy$ be "small." We find the probability that $Y$ lies between $y$ and $y+dy$.
This will be approximately $f_Y(y)\,dy$.
Neglecting terms in higher powers of $dy$, the probability that the second largest lies between $y$ and $y+dy$ is the probability that some $X_i$ lies in this interval times the probability that $n-2$ of the $X_i$ lie below $y$ and $1$ lies above $y+dy$.
The $X_i$ that lies between $y$ and $y+dy$ can be chosen in $n$ ways. The probability it lies in the interval is approximately $F(y)\,dy$.
The probability that $n-2$ of the remaining $X_i$ lie below $y$, and $1$ lies above, is $\binom{n-1}{1}(1-F(y))^1 (F(y))^{n-2}$. "Thus,"
$$f_Y(y)=n\binom{n-1}{1}f(y)(1-F(y))(F(y))^{n-1}.$$
For the cdf $F_Y(y)$, integrate from $-\infty$ to $y$. The integral is in principle easy, make the substitution $u=F(y)$.
A: A different way to show André Nicolas' result:
Let $Y$ be the second largest $X_i$. Then $Y \le x$ if and only if there is at most one $i$ such that $X_i > x$. Let $F(x) = \mathbb{P}(X_1 \le x)$ be the distribution function of $X_1.$
Using the fact that the $X_i$ are i.i.d.:
$$
\begin{align}
\mathbb{P}(Y \le x) &= \mathbb{P}(\{\exists! i: X_i > x\}) + \mathbb{P}(\{\forall i: X_i \le x\})\\
&= \sum_i \mathbb{P}(X_1 \le x, \dots, X_{i-1} \le x, X_i > x, X_{i+1} \le x, \dots, X_n \le x) + \mathbb{P}(X_1 \le x, \dots, X_n \le x) \\
&= \sum_i \mathbb{P}(X_i > x) \prod_{j \neq i} \mathbb{P}(X_j \le x) + \prod_i \mathbb{P}(X_i \le x) \\
&= \sum_i (1-F(x)) F(x)^{n-1} + F(x)^n \\
&= n (1-F(x)) F(x)^{n-1} + F(x)^n.
\end{align}
$$
Differentiate to get the density $f_Y$.

P.S.: One can also show $\mathbb{P}(Z > x, Y \le x) = \mathbb{P}(Y \le x) - \mathbb{P}(Z \le x)$, even if the $X_i$ are not i.i.d.
A: What is the probability that $N = 2$? i.e. what is the probability that $X_1 > X_2$? Well since $X_1$ and $X_2$ are iid, $\mathbb P(X_1 > X_2) = \mathbb P(X_2 < X_1)$, and by continuity $\mathbb P(X_1 = X_2) = 0$. Therefore $\mathbb P(N = 2) = \mathbb P(X_1 > X_2) = 1/2$. 
What is the probability that $N = 3$? If $N=3$, then in particular $N \neq 2$, so $X_1 < X_2$. But $N = 3$ means $X_3$ comes second, so we must have $X_1 < X_3 < X_2$. Since this is one of the $6$ possible configurations of $\{1,2,3\}$, $\mathbb P(N=3) = 1/6$. 
What is the probability that $N=4$? Since $N \neq 2$ and $N \neq 3$, we must have $X_1 < X_2$, and we cannot have $X_1 < X_3 < X_2$, so we either have $X_3 < X_1 < X_2$ or $X_1 < X_2 < X_3$. But then $X_4$ comes in second so that leaves us with two possible permutations of $\{1,2,3,4\}$ ; $\mathbb P(N=4) = \frac 2{24} = \frac 1{12}$.
The probability that $N = k$, $k \ge 2$ will be given as follows : you consider all the possibilities $N \neq 2,3,\dots,k-1$ and then $N = k$. The steps $2,3,\dots,k-1$ will each give $1,2,\dots,k-2$ independent choices, and $N=k$ will force the position of $X_k$ to be second. You will then compute the probability of obtaining one of the $(k-2)!$ permutations of the possible orderings of $\{1,2,\dots,k\}$, i.e. $\mathbb P(N=k) = \frac{(k-2)!}{k!} = \frac 1{k(k-1)}$.
I didn't just give you the solution, I explained to you how I thought about it ; this is how you should've come up with the solution. Explore small cases, find a pattern. =D
Hope that helps,
