Calculating the Maximum Likelihood Estimate I'm trying to understand MLE, but struggling with a question/example, so was hoping for an explanation if someone can help.
There's a ring toss game which is repeated until it's succeeds for the first time. Each time the game is played, it succeeds with the probability p; and each attempt is independent.
I have the following probability function of X.
$\ f(x)=p(1−p)^{x-1}$
I'm trying to work out the MLE of p (given a single observation x of X). Would someone know how to go about this?
What I'm struggling with here is that the number of attempts, as it isn't fixed and instead is just 'n'.
 A: Your likelihood function is
$$
L(p \mid x) = p \left(1 - p\right)^{x - 1}.
$$
Differentiating with respect to $p$,
$$
\begin{aligned}
\frac{d L}{d p}(p \mid x)
&= \left(1 - p\right)^{x - 1} - p (x - 1) \left(1 - p\right)^{x - 2} \\
&= (1 - p)\left(1 - p\right)^{x-2} -  (p x - p) \left(1 - p\right)^{x - 2} \\
&=  (1 - p - p x + p) \left(1 - p\right)^{x - 2} \\
&=  (1 - p x) \left(1 - p\right)^{x - 2}.
\end{aligned}
$$
Setting this equal to zero (since the derivative of $L(p \mid x)$ at $p$ is zero if $L(p \mid x)$ is at a maximum), we get
$$
(1 - p x) \left(1 - p\right)^{x - 2} = 0.
$$
Assuming $p < 1$, it follows that $\left(1 - p\right)^{x - 2} \neq 0$, so $1 - p x = 0$, and hence $p = 1 / x$.
Thus, the maximum likelihood estimate of $p$ is
$$
\hat{p} = \frac{1}{x}.
$$
A: Since you have a single observation, you have $n=1$.  So the likelihood function for the observed data is:
$$L_x(p) = p(1-p)^{x-1} \quad \quad \quad \text{for all } 0 \leqslant p \leqslant 1.$$
As the name suggests, the maximum-likelihood estimator is the value that maximises the likelihood function:
$$\hat{p}(x) = \underset{0 \leqslant p \leqslant 1}{\text{arg max }} L_x(p)$$
In this particular case, this is just a basic calculus problem, where you are trying to maximise the likelihood function with respect to a continuous argument variable.  Since the likelihood function is a product of non-negative parts that depend on $p$, it is usual to find the maximising value by maximising the log-likelihood function (i.e., the logarithm of the likelihood function).  If you apply standard calculus methods to maximise this function, you will be able to derive the maximum-likelihood estimator $\hat{p}$.
