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

\begin{align} P(M_{n}=a, M_{n+1}=b)= P(M_{n+1}=b \mid M_n=a)P(M_n=a) \end{align}

Either $M_{n+1} = M_{n}$, which is equivalent to the event $X_{n+1} < M_n$, or $M_{n+1} = M_n$, which is equivalent to $X_{n+1} > M_n$. So the joint PDF should be

\begin{align} P(M_{n}=a, M_{n+1}=a) &= P(M_n=a)P(X_{n+1}<a) \\ &= n\binom{n-1}{j-1}\; f_x(a)\;F_X(a)^{\color{red}{j}} \;(1-F_X(a))^{n-j} \end{align} \begin{align} P(M_{n}=a, M_{n+1}=x_{n+1}) &= P(M_n=a)P(X_{n+1}>a) \\ &= n\binom{n-1}{j-1} \; f_x(a) \; F_X(a)^{j-1} \; (1-F_X(a))^{\color{red}{n-j+1}} \end{align}

Is this right?

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The $X_i$ are independent, so we have:

$$ \begin{eqnarray*} P\left(M_n>m_n\,,\,M_{n+1}<m_{n+1}\right)&{}={}&P\left(M_n>m_n\,,\,M_n<m_{n+1}\,,\,X_{n+1}<m_{n+1}\right) \newline &&\newline &{}={}&P\left(m_{n+1}>M_n>m_n\right)P\left(X_{n+1}<m_{n+1}\right) \newline &&\newline &{}={}&\bigg(\left(1-F(m_n)\right)^n{}-{}\left(1-F(m_{n+1})\right)^n\bigg)F(m_{n+1}){\textbf{1}}_{\left\{m_n<m_{n+1}\right\}}\,. \end{eqnarray*} $$ This defines the joint distribution.

Note : Since we are not told that the $\,X_i\,$ are discrete random variables, I don't think your proposed approach is sufficiently general.

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