Find the Maximum Likelihood Estimator for the Given Equation

Question

How should I go about determining the maximum likelihood estimator for beta (part B)? Also, what is the distribution this question is asking for, and how can I calculate the probability that the parameter is within 0.05 of 3 (parts c and d)?

Answer 1

The likelihood for the sample is $$\prod_{i=1}^n\frac{\log 2}{\beta}e^{-\frac{\log 2}{\beta}x_i}=\frac{(\log 2)^{n}}{\beta^n}\exp\left\{{-\frac{\log 2}{\beta}\sum_{i=1}^n x_i}\right\}$$

so the logarithm of the likelihood is $$n\log(\log 2)-n\log \beta-\frac{\log 2}{\beta}\sum_i x_i$$

To find the MLE of $\beta$, simply take the derivative and set to $0$:

$$-\frac{n}{\beta}+\frac{\log 2}{\beta^2}\sum _i x_i=0\\ \rightarrow\hat \beta=\frac{\log 2}{n}\sum_i x_i$$

You can verify that the function is increasing on the left by plugging in $\frac{\log 2}{n+1}\sum_i x_i$ to get the derivative to be $-n(n+1)+(n+1)^2>0$. Thus $\hat \beta$ is a global maximum.

The central limit theorem says that when you add things together they become normal. First begin by finding $E(x_i)=\beta/\log 2$ and $V(x_i)=\beta^2/(\log 2)^2$ so that $E(\hat \beta)=(\log 2) / n * n\beta/\log 2=\beta$ and $V(\hat \beta)=(\log 2)^2/n^2 * n \beta^2/(\log 2)^2=\beta^2/n$ so that $\hat \beta\sim \mathscr N(\beta, \beta^2/n)$.

With $\beta=3$, we have $|\hat\beta-3|\le .05$ is $|Z| \le .05 / .3=1/6$ with $Z$ having standard normal distribution ($\mathscr N(0, 1)$) which is $.1325$ (entered into the statistical software known as "R").

The average number of samples would then be $7*.1325$