Question

Let \( x_1 = 1.1, x_2 = 2.2, x_3 = 3.3 \) be the observed values of a random sample of size three from a distribution with the probability density function \[ f(x; \theta) = \begin{cases} \dfrac{1}{\theta} e^{-x/\theta}, & x > 0, \\ 0, & \text{otherwise}, \end{cases} \] where \( \theta \in \Theta = \{ 1, 2, \dots \} \) is the unknown parameter. Then the maximum likelihood estimate of \( \theta \) equals ......... 
 

Hint:

The maximum likelihood estimate for exponential distributions is the sample mean.

Answer 1

Correct Answer: 1.9 - 2.1

The pdf is
\[f(x;\theta)=\frac{1}{\theta}e^{-x/\theta}, \quad x>0,\qquad \theta\in\Theta={1,2,\dots}.\]

Likelihood

For observations \(x_1=1.1,\ x_2=2.2,\ x_3=3.3\),
\[L(\theta)=\prod_{i=1}^3 \frac{1}{\theta}e^{-x_i/\theta} = \theta^{-3}\exp!\left(-\frac{\sum x_i}{\theta}\right),\]
where \(\sum x_i = 6.6\).

The log-likelihood is
\[\ell(\theta)=-3\ln\theta-\frac{6.6}{\theta}.\]

Continuous maximizer

Differentiating,
\[\ell'(\theta)=-\frac{3}{\theta}+\frac{6.6}{\theta^2}=0 \Rightarrow \theta=\frac{6.6}{3}=2.2.\]

Discrete parameter space

Since \(\theta\in{1,2,\dots}\), the MLE is the integer nearest to (2.2), namely
\[\hat\theta=2.\]

\[\boxed{\hat\theta=2}\]