Let X = X1, X2, . . . , Xn be i.i.d. random variables from a population with a density if x ≥ θ f(x;θ) = 2θ^2/x^3, if x ≥ θ, 0 elsewhere. What is the Maximum Likelihood Estimator of θ? Is it unbiased?
So far calculated the log likelihood to be nlog2 +2nlogθ - 3nlogx. Deriving this and setting to zero does not make sense to me?
Thanks for any assistance
Using proper notation would help make the computation more understandable.
The density can be expressed with the use of an indicator function: $$f(x \mid \theta) = \frac{2\theta^2}{x^3} \mathbb{1}(x \ge \theta),$$ where $$\mathbb{1}(x \ge \theta) = \begin{cases}1, & x \ge \theta \\ 0, & x < \theta. \end{cases}$$ Now the joint density of the sample $\boldsymbol x = (x_1, x_2, \ldots, x_n)$ is simply $$f(\boldsymbol x \mid \theta) = \prod_{i=1}^n f(x_i \mid \theta) = (2\theta^2)^n \prod_{i=1}^n x_i^{-3} \mathbb{1}(x_i \ge \theta).$$ But observe that the product $$\prod_{i=1}^n \mathbb{1}(x_i \ge \theta) = \mathbb{1}(\min_i x_i \ge \theta) = \mathbb{1}(x_{(1)} \ge \theta);$$ that is to say, the product of the indicator functions is $1$ if and only if each observation $x_i$ is at least $\theta$, or equivalently, the smallest $x_i$ is at least $\theta$; this is more compactly written as the first order statistic $x_{(1)} = \min_i x_i$. Therefore, a likelihood function for $\theta$ given the sample $\boldsymbol x$ is given by $$L(\theta \mid \boldsymbol x) = \theta^{2n} \mathbb{1}(x_{(1)} \ge \theta).$$ We need not retain the factors $2^n$ nor $\prod x_i^{-3}$, because these are constant with respect to $\theta$ for a given fixed set of observations and sample size $n$.
The only remaining thing to do is to determine, for a fixed set of observations $\boldsymbol x$, what value of $\theta$ maximizes $L$. Clearly, as $\theta$ increases, the factor $\theta^{2n}$ also increases (since $n$ is a positive integer). But its value is strictly limited by the requirement that $\theta$ must not exceed the smallest observation $x_{(1)}$; otherwise, the value of the indicator function is zero and the likelihood of having observed our sample is also zero.
Now, suppose $\hat\theta$ is the maximum likelihood estimator based on the above reasoning. Determine the distribution of this estimator for a fixed but unknown true value of the parameter $\theta$, and compute its expected value as a function of $\theta$. If $\mathrm{E}[\hat\theta] = \theta$, then $\hat \theta$ is unbiased. But we can already intuitively sense that $\hat\theta$ is necessarily biased, because our estimator can never be an underestimate of the true value of the parameter from which the sample was drawn: for instance, if our observed sample was $\boldsymbol x = (3, 2, 3, 5, 10)$, we immediately know without doing any computation that $\theta \le \hat\theta = 2$. We can never obtain a $\hat\theta$ that is smaller than the true parameter $\theta$, so its expectation cannot equal $\theta$.