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
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 $\theta \ge 4$.