Let $x_1 =1.1, x_2 =2.2 \space \text{and} \space x_3 =3.3$ be the observed values of a random variable of sample size of 3 from a distribution with probability density function
$$ f(x,\theta) =\begin{cases}
\frac{1}{\theta}e^{\frac{-x}{\theta}} & x \ge 0 \\
0 & \text{otherwise} \\
\end{cases} $$
where $\theta \in \{1,2,3…\}$ is the unknown parameter. The maximum likelihood estimate of $\theta$ is _____________
Wht i did
The likelihood function is
$$ L(\theta) = \prod_{i=1} ^{n} f(x_i,\theta) $$
$$ \Bigg(\frac{1}{\theta} \Bigg) ^n e^{\frac{{-\sum x_i}}{\theta}}$$
Taking Log on both Side
$$ \ln(L(\theta)) = n\ln(\frac{1}{\theta}) – \frac{{\sum x_i}}{\theta} $$
Now we differentiate wrt $\theta$
$$\frac{d \ln(L(\theta))}{d\theta} = n\theta + \frac{{\sum x_i}}{\theta^2} $$
equating it to $0$
$$ \theta^3 = – \frac{{\sum x_i}}{n} $$
When I substitute the values for $x$ and $n$ I get $1.3$ . But it not the correct as per answer Key.
Is my method wrong, if so is there any other way to do this problem.
Thank You
Best Answer
The likelihood is $$L(\theta) = \theta^{-n} \exp \left(- \frac{1}{\theta} \sum x_i \right). \tag{1}$$ Your calculation of the derivative of the log-likelihood is incorrect; it should be $$\frac{d}{d\theta}[\log L(\theta)] = -n\theta + \frac{\sum x_i}{\theta^2}.$$ This yields the unrestricted MLE $$\hat \theta = \frac{\sum x_i}{n},$$ and if $\theta$ were not necessarily an integer, it would suggest $\hat \theta = 2.2$. But since we require $\theta$ to be a positive integer, we simply substitute into Equation $(1)$: $$L(1) = \exp(-6.6) \approx 0.00136037, \\ L(2) = 2^{-3} \exp(-3.3) \approx 0.0046104, \\ L(3) = 3^{-3} \exp(-2.2) \approx 0.00410382.$$ We then see $L(2)$ is maximal for the restricted likelihood. The purpose of computing the unrestricted likelihood is to show that the local extremum at $\hat \theta$ implies that $L'(\theta) < 0$ when $\theta > \hat \theta$, hence we need not consider $L(\theta)$ for $\theta \ge 4$.