i.i.d random variables

Question

Looking at this question I. i. d distributions as best car offers I wonder about the following one:

Can we find the distribution function of $X_N$ where,

$$ N= \min \{ n \geq 1 \mid X_n > X_0 \} $$

where the $X_i$ are independent identically distributed random variables with say, density function $f$ and distribution function $F$ ?

Any ideas will be appreciated!

edit: I have edit the question. I forgot to type the $X_N$.

In the answer in the link there is the probability $ P[N=n]$ expressed with respect to the expectation. How can we derive a formula with involving the expectation and depends only on $f$ and $F$ ?

P.S. I sorry if I write something silly, I have just start studying these things.

Thank you!

Answer 1

This might be one of the rare cases I know where the CDF is the most convenient approach... For every $n\geqslant0$, let $M_n=\max_{0\leqslant k\leqslant n}X_k$, then, for every $x$, $$ [X_N\leqslant x]=\bigcup_{n\geqslant1}[x\geqslant X_n\gt X_0\geqslant M_{n-1}]=\bigcup_{n\geqslant1}[x\geqslant M_n,A_n], $$ where $$ A_n=[X_n\gt X_0\geqslant M_{n-1}]. $$ The event $A_n$ is concerned with the ordering of the random variables $(X_k)_{0\leqslant k\leqslant n}$ and $M_n$ is their maximum hence $M_n$ and $A_n$ are independent. And $A_n$ is realized when the two largest values in an i.i.d. sample of size $n+1$ are obtained at times $n$ and $0$, hence, assuming from now on that the CDF $F$ is continuous, $$ P(A_n)=\frac1{n(n+1)}. $$ Thus, $$ P(X_N\leqslant x)=\sum_{n\geqslant1}P(M_n\leqslant x)P(A_n)=\sum_{n\geqslant1}\frac{F(x)^{n+1}}{n(n+1)}. $$ Differentiating both sides yields the PDF $g$ of $X_N$ as $$ g(x)=\sum_{n\geqslant1}f(x)\frac{F(x)^{n}}{n}=-f(x)\log(1-F(x)). $$ Finally, the CDF $G$ of $X_N$ is such that $$ G(x)=F(x)+(1-F(x))\log(1-F(x)). $$

Answer 2

I do not know whether it is elegant or not, but what I'd try is as follows: first, $$ \mathbb{P}\{N=1\} = \mathbb{P}\{X_1 > X_0\} = \int_{\mathbb{R}} \mathbb{P}\{X_1 > x,\ X_0 \in dx\} = \int_{\mathbb{R}} dx f(x)\left(1-F(x)\right) $$ the last step by independence. Then, \begin{align*} \mathbb{P}\{N=2\} &= \mathbb{P}\{X_2 > X_0,\ X_1 \leq X_0\} = \int_{\mathbb{R}} \mathbb{P}\{X_0\in dx\}\mathbb{P}\{X_2 > x,\ X_1 \leq x\} \\ &=\int_{\mathbb{R}} \mathbb{P}\{X_0\in dx\}\mathbb{P}\{X_2 > x\}\mathbb{P}\{ X_1 \leq x\} = \int_{\mathbb{R}} dx f(x)(1-F(x))F(x) \end{align*} and similarly, more generally \begin{align*} \mathbb{P}\{N=k\} &= \int_{\mathbb{R}} dx f(x)(1-F(x))F(x)^{k-1}. \end{align*} Does that look/sound correct?