Order Statistics w/ Maximum 
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? 
 A: 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.  





*

*In $\frac{1}{11}$ cases, the $11^\text{th}$ term will be LARGER than $M_{10}$, in which case $M_{11} > M_{10}$. 


The following diagram plots the exact theoretical solution $g(x_{(10)}, x_{(11)})$ derived above:

Finally, the next diagram compares the empirical joint pdf of $(M_{10}, M_{11})$  to the theoretical solution $g(x_{(10)}, x_{(11)})$ derived above:

