Maximum Likelihood Estimator of Bin(m,p)

[Ted123]

Homework Statement

Let $\displaystyle X_1 ,..., X_n \stackrel {\text{i.i.d.}}{\sim} \text{Bin} (m,p)$ where $m$ is known. Find the MLE $\hat{p}_n$ of $p$ and hence show that $\hat{p}_n$ is unbiased.

The Attempt at a Solution

Can anyone check my attempt please?

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

$\displaystyle \log (L) = \log \left[ \sum_{i=1}^n {m \choose x_i} \right] + \log \left( p^{\sum_{i=1}^n x_i} \right) + \log \left[ (1-p)^{\sum_{i=1}^n (m-x_i)} \right]$

$\displaystyle = \prod_{i=1}^n \log \left[ {m \choose x_i} \right] + \log (p)\sum_{i=1}^n x_i + \log(1-p) \sum_{i=1}^n (m-x_i})$

$\displaystyle \frac{\partial}{\partial p} \log(L) = \frac{1}{p} \sum_{i=1}^n x_i - \frac{1}{1-p} \sum_{i=1}^n (m-x_i)$

Setting $\displaystyle \frac{\partial}{\partial p} \log(L) = 0$

$\displaystyle \frac{1}{p} \sum_{i=1}^n - \frac{1}{1-p} \sum_{i=1}^n (m-x_i) =0$

$\displaystyle (1-p) \sum_{i=1}^n x_i - p \sum_{i=1}^n (m-x_i) =0$

$\displaystyle p = \frac{\sum_{i=1}^n x_i}{\sum_{i=1}^n x_i - \sum_{i=1}^n (m-x_i)}$

The denominator reduces to $nm$ and we have

$\displaystyle p = \frac{\sum_{i=1}^n x_i}{nm} = \frac{\overline{x}}{m}$

$\displaystyle \hat{p}_n = \frac{\overline{X}}{m}$

$\hat{p}_n$ is unbiased since $\displaystyle \mathbb{E}[\hat{p}_n] = \mathbb{E}\left[ \frac{\overline{X}}{m}\right] = \frac{1}{m} \mathbb{E}\left[\:\overline{X}\: \right] = \frac{1}{m} \mathbb{E} \left[ \frac{1}{n} \sum_{i=1}^n x_i \right] = \frac{1}{nm} \sum_{i=1}^n \mathbb{E}[X_1] = \frac{n}{nm} (mp) = p$

$\therefore \hat{p}_n$ is an unbiased estimator of $p$ .

 

[mighty2000]

Your attempt seems to be correct. You have correctly derived the MLE for p and showed that it is unbiased. Your explanation is clear and concise. Good job! If you have any further questions, feel free to ask.