log likelihood function and MLE for binomial sample

Question

Let $X_1,X_2,...;X_n$ be a random sample with $X_i$~$Binomial(m,p)$ for $i=1,...,n$ and $m=1,2,3,...$ and let $p\in (0,1)$. We assume $m$ is known and we are given the following data $x_1,...,x_n\in\{0,...,m\}$

Write up the log-likelihood function and find the MLE $\hat{P}ML$ for p

I'm not quite sure how to approach this. This is what I've tried:

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I believe the likelihood function of a Binomial trial is given by

$P_{X_i}(x;m)=$ ${m}\choose{x} $$p^x(1-p)^{m-x}$

From here I'm kind of stuck. I'm uncertain how I find/calculate the log likelihood function.

I've understood the MLE as being taking the derivative with respect to m, setting the equation equal to zero and isolating m (like with most maximization problems). So finding the log likelihood function seems to be my problem

Edit: I might be misunderstanding it but could the log likelihood function simple be log of the likelihood function? so $log(P_{X_i}(x;m))$

Answer 1

The likelihood is

$$L(p)=\prod_{i=1}^nP_p(X=x_i)=\prod_{i=1}^{n}{m\choose x_i}p^{x_i}(1-p)^{m-x_i}$$

The log-likelihood is thus

$$\log L(p)=\log\left(\prod_{i=1}^{n}{m\choose x_i}\right)+\log(p)\sum_{i=1}^nx_i+\log(1-p)\left(nm-\sum_{i=1}^nx_i\right)$$

Let $M=\log\left(\prod_{i=1}^{n}{m\choose x_i}\right)$ (which does not depend on $p$):

$$\log L(p)=M+\log(p)\sum_{i=1}^nx_i+\log(1-p)\left(nm-\sum_{i=1}^nx_i\right)$$

$$\log L(p)=M+n\log(p)\bar x+n\log(1-p)(m-\bar x)$$

$$\dfrac{\partial\log L}{\partial p}=\frac{n\bar x}{p}-\frac{n(m-\bar x)}{1-p}$$

This last expression is zero if

$$\frac{\bar x}{p}=\frac{m-\bar x}{1-p}$$

$$\bar x-\bar xp=mp-\bar xp$$

Hence the ML estimator is:

$$\hat p=\dfrac{\bar x}{m}$$

Answer 2

I think you are correct, but usually you take the product of P_X over all observations X_1,...,X_n to compute the Likelihood of a sample. This translates to the sum of log probabilities.

I think you are interested in finding the parameter p that maximises the (Log)Likelihood rather than the known quantity m.