Order Statistics w/ Maximum

Question

Let $X_1,X_2,\dotsc$ be i.i.d. r.v.s with CDF $F$, and let $M_n = \max(X_1,X_2, \dotsc,X_n)$. Find the joint distribution of M$_n$ and M$_{n+1}$, for each n ≥ 1.

So, CDF of $M_{n+1}$ is given as, $P(M_{n+1} \le x) = P(X_1 \le x, X_2 \le x, \dotsc, X_n \le x, X_{n+1} \le x ) = F(x) * F(x) * ....(n+1) \times = F(x)^{n+1}$.

We want to consider two cases: $P(M_n \le a,M_{n+1} \le b)$. However, after getting these facts, I am lost. Can someone help me in finding this joint distribution?

You need to require a<=b since otherwise the probability is 0 You should know P(Mn <=a, Mn+1<=b) = X$_1$,<=a,...X$_n$<=a, Xn+1<=b)=F(a)$^n$ F(b) — Michael R. Chernick, Nov 10 '17 at 05:39
Related: https://stats.stackexchange.com/questions/164222/joint-probability-distribution-of-k-le-n-order-statistics — kjetil b halvorsen, Nov 10 '17 at 10:08

wolfies · Answer 1 · 2017-11-11T12:26:27.493

Question:

Let:

$(X_1, X_2, \dots, X_n, X_{n+1})$ denote a sample of size $(n+1)$ on random variable $X$ with pdf $f(x)$
$M_n$ denote the maximum of the first $n$ values, and $M_{n+1}$ the sample max of all $n+1$ values
$g(x_{(n+1)})$ denote the pdf of the sample maximum, given a sample of size $n+1$.
$g(x_{(n)},x_{(n+1)})$ denote the joint pdf of the $n^{\text{th}}$ and $(n+1)^{\text{th}}$ order statistics in a sample of size $n+1$ (i.e. the joint pdf of the second largest and largest values).

The pdf of $g$ and pdf of $g(.,.)$ are textbook material, and can be expressed as a function of $f(x)$.

The question is to find the joint pdf of $(M_n, M_{n+1})$.

Answer: There are two possibilities:

1) $X_{n+1} \leq M_n$, in which case $M_{n+1}= M_n$. This happens with probability $\frac{n}{n+1}$.

or

2) $X_{n+1} > M_n$, in which case $M_{n+1} > M_n$. This happens with probability $\frac{1}{n+1}$.

Accordingly, the joint pdf of $(M_n, M_{n+1})$ can be written:

$$\text{pdf}(M_n, M_{n+1}) = \left\{ \begin{array}{cc} \frac{n}{n+1} g(x_{(n+1)}) \;\;\;\; & \text{if } M_{n+1} = M_{n}\\ \frac{1}{n+1} g(x_{(n)}, x_{(n+1)}) & \text{if } M_{n+1} > M_{n} \\ \end{array} \right. $$

Example (with simulation check)

Let the parent random variable $X$ be standard Normal with pdf $f(x)$:

Then, the pdf of the sample maximum, in a sample of size $n+1$, is $g(x_{(n+1)})$:

where I am using the OrderStat function from the mathStatica package for Mathematica to automate,

and the joint pdf of the $n^{\text{th}}$ and $(n+1)^{\text{th}}$ order statistics in a sample of size $n+1$ is $g(x_{(n)}, x_{(n+1)})$:

Simulation check

The following Monte Carlo check generates 10 million samples, each of size $n = 10$. For each case, it then generates an $11^\text{th}$ term, and compares it to the first 10 values.

In $\frac{10}{11}$ cases, the $11^\text{th}$ term will be smaller than $M_{10}$, in which case $M_{11} = M_{10}$. The following diagram compares the empirical pdf of $M_{11}$ (blue) to the theoretical solution $g(x_{(11)})$ derived above (red dashed). They are so close, it is difficult to see any difference.