Problem source

Write down the probability of obtaining $k$ heads in $n$ tosses of a fair coin. Now suppose that $k$ is known but $n$ is unknown. A \textit{maximum likelihood estimator} (MLE) of $n$ is defined to be a value (which must be an integer) of $n$ which maximizes the probability of $k$ heads.
A friend has thrown a fair coin a number of times. She tells you that she has observed one head. Show that in this case there are \textit{two} MLEs of the number of tosses she has made. She now tells you that in a repeat of the exercise she has observed $k$ heads.  Find the two MLEs of the number of tosses she has made.  She next uses a coin biased with probability $p$ (known) of showing a head, and again tells you that she has observed $k$ heads. Find the MLEs of the number of tosses made. What is the condition for the MLE to be unique?

Solution source

\begin{align*}
&& \mathbb{P}(k \text{ heads} |  n\text{ tosses}) &= \binom{n}k 2^{-n} \\
&& \mathbb{P}(1 \text{ head} |  n\text{ tosses}) &= n2^{-n} \\
\Rightarrow && \frac{ \mathbb{P}(1 \text{ head} |  n+1\text{ tosses}) }{ \mathbb{P}(1 \text{ head} |  n\text{ tosses}) } &= \frac{n+1}{2n}
\end{align*}

Which is less than $1$ unless $n \geq 1$. Therefore the MLE is $n = 1$ or $n= 2$.

\begin{align*}
 \frac{ \mathbb{P}(k \text{ head} |  n+1\text{ tosses}) }{ \mathbb{P}(k \text{ head} |  n\text{ tosses}) } &= \frac{\binom{n+1}{k}}{2 \binom{n}{k}} \\
&= \frac{(n+1)!(n-k)!}{2n!(n+1-k)!} \\
&= \frac{n+1}{2(n+1-k)}
\end{align*}

This is less than or equal to $1$ if $n+1 = 2(n+1-k) \Leftrightarrow n= 2k-1$, therefore the MLEs are $2k-1$ and $2k$.

If the coin is biased, we have

\begin{align*}
&&  \frac{ \mathbb{P}(k \text{ head} |  n+1\text{ tosses}) }{ \mathbb{P}(k \text{ head} |  n\text{ tosses}) } &= \frac{\binom{n+1}{k}p^kq^{n+1-k}}{\binom{n}{k}p^kq^{n-k}} \\
&&&= \frac{n+1}{(n+1-k)}q \\ 
\\
&& 1 & \geq \frac{n+1}{(n+1-k)}q \\
\Leftrightarrow && (n+1)(1-q) &\geq k \\
\Leftrightarrow && n+1 & \geq \frac{k}{p}
\end{align*}

Therefore the probability is increasing until $n+1 \geq \frac{k}{p}$. If $\frac{k}p$ is an integer the MLEs are $\frac{k}{p}-1$ and $\frac{k}p$, otherwise it is $\lfloor \frac{k}{p} \rfloor$ and the MLE is unique.