Let be $X_1, X_2\dots X_n$ independent binomially distributed random variables with probability $p$ and length $m$. We denote $P(X_i=k)$, the probability of $k$ sucesses, by $b(k;m,p)$. Both parameters $p$ and $m$ are unknown. We know that $X_{(n)}=\max(X_1,X_2,...,X_n)$ is a consistent estimator for length $m$ (see https://math.stackexchange.com/a/4618523/579544).
Is $X_{(n)}$ unbiased?
My approach:
For sake of simplicity I assume that $n=3$ and $X_{(3)}=\max(X_1,X_2,X_3)=k$. By inclusion-exclusion principle we see that
\begin{align*} &P(X_{(n)}=k)=P(\{X_1=k\}\cup \{X_2=k\}\cup \{X_3=k\})\\ &=P(X_1=k)+P(X_2=k)+P(X_3=k)\\ &-P(\{X_1=k\}\cap \{X_2=k\})-P(\{X_1=k\}\cap \{X_3=k\})-P(\{X_2=k\}\cap \{X_3=k\})\\ &+P\left(\{X_1=k\}\cap \{X_2=k\}\cap \{X_3=k\}\right)\\ &=3\cdot b(k;m,p)-3\cdot b(k;m,p)^2+b(k;m,p)^3. \end{align*} So the expected value $\mathbb{E}(X_{(n)})$ can be calculated by \begin{align*} \mathbb{E}(X_{(n)})=\sum\limits_{m=k}^{\infty}3m\cdot b(k;m,p)-3m\cdot b(k;m,p)^2+m b(k;m,p)^3. \end{align*}
I don't know how to proceed/manipulate the series to get any further results? Is this a valid approach at all? Or is there a different/easier one? Maybe the expected value doesn't exist at all?
Observe that \begin{align*} \mathbb E[X_{(n)}] &= \sum_{k=1}^m \mathbb P[X_{(n)} \geq k]\\ &=\sum_{k=1}^m (1-\mathbb P[X_1 < k]^n)\\ &=m-\sum_{k=1}^m \mathbb P[X_1 < k]^n\\ &<m \end{align*} Therefore $X_{(n)}$ is biased.
The first equation is due to the fact that for any positive discrete random variable $Y$, \begin{align*} \mathbb E[Y] &= \sum_{i=0}^\infty i \mathbb P[Y=i]\\ &= \sum_{i=0}^\infty \sum_{k=1}^i \mathbb P[Y=i]\\ &= \sum_{k=1}^\infty \sum_{i=k}^\infty \mathbb P[Y=i]\\ &= \sum_{k=1}^\infty \mathbb P[Y\geq k] \end{align*}