# Determine the MLE of $\theta$

Let $$\{x_1, x_2, \cdots, x_n \}$$ be an observed sample from a distribution with probability density function

$$f(x) = \begin{cases}\frac{x^3}{2 \theta^3}e^{-\theta x} & x > 0 \\0 & \textrm{otherwise}\end{cases}$$ where $$\theta > 0$$ is unknown.

(a) Determine the MLE of $$\theta$$

(b) Determine the MLE of $$\psi(\theta) = \textrm{Var}(X_1)$$

Attempt

(a)

$$(x_1, x_2, \cdots, x_n) \sim \textrm{Gamma}(2, 1)$$

$$L(\theta | \bar{x}) = \prod_{i=1}^{n} \frac{x_i^3}{2 \theta^3}e^{-\theta x_i} = \left(\frac{1}{2 \theta^3}\right)^n \prod_{i=1}^{n} x_i^3 e^{-\theta x_i} = \left(\frac{1}{2 \theta^3}\right)^n \bar{x}^{3} e^{-\theta\sum_{i=1}^{n} x_i} \\ = \left(\frac{1}{2 \theta^3}\right)^n \bar{x}^{3} e^{-\theta \bar{x}}$$

$$L(\theta | \bar{x}) = \ln L = n \ln (\frac{1}{2 \theta^3}) + 3 \ln(\bar{x}) - \theta \bar{x} \ln (e)$$

Derivative:

$$0 = \frac{3}{\theta} + 0 -\bar{x}$$

so $$\theta = -\frac{3}{\bar{x}}$$

now to see if it's derivative is less than 0

$$S(\theta | \bar{x}) = \frac{-3}{\theta} - \bar{x}$$

$$S'(\theta | \bar{x}) = \frac{3}{\theta^2} < 0$$

therefore $$\theta = -\frac{3}{\bar{x}}$$ is the Maximum likelihood estimate (MLE)

not sure how to do b

• You have made some mistakes in your Likelihood equation. Mar 7, 2019 at 5:50
• what is wrong with it? Mar 7, 2019 at 6:00
• nvm I fixed it. Mar 7, 2019 at 6:07
• There are still errors in your likelihood. Mar 7, 2019 at 6:12
• whats wrong with mine ? Mar 7, 2019 at 6:16

Your likelihood is wrong. The correct likelihood is, $$L(\theta | \bar{x}) = \prod_{i=1}^{n} \frac{x_i^3}{2 \theta^3}e^{-\theta x_i} = \left(\frac{1}{2 \theta^3}\right)^n \prod_{i=1}^{n} x_i^3 e^{-\theta x_i} = \left(\frac{1}{2 \theta^3}\right)^n \left(\prod_{i=1}^{n} x_i^3\right) e^{-\theta\sum_{i=1}^{n} x_i}$$
The log-likelihood is then given by, $$\log L(\theta | \bar{x}) = -n\log2 -3n \log \theta + 3 \sum_{i = 1}^n \log x_i -\theta\sum_{i=1}^{n} x_i$$
Edit: By the way, I also see that you wrote $$\textrm{Gamma(2, 1)}$$ but the pdf expression that you wrote doesn't match what I see on this wikipedia link.