The MLE of this weird density function

Question

I'm trying to solve the first question of this problem set. I'm a beginner in this subject and it's the very first question I'm trying to solve. I really need help.

The question is:

Let $X_1,\ldots,X_n$ be $n$ i.i.d. random variables with density $f_{\theta}$ with respect to the Lebesgue measure. For each case below find the MLE of $\theta$.

$$f_{\theta}(x)= \theta\tau^{\theta}x^{−(\theta+1)}\unicode{x1D7D9}_{x \ge\tau},\ \theta>0$$

where $\tau>0$ is a known constant.

I supposed the function is concave and I took the derivative of the log and I equaled to zero:

$$\hat\theta = \frac{n}{\log(\frac{x_1\cdots x_n}{\tau^n})}$$ provided that $x_1,\ldots, x_n$ are greater or equal than $\tau$. The MLE is zero otherwise.

am I right?

Answer 1

It's ok.

With different notation I get

$$\hat{\theta}_\text{ML}=\frac{n}{\sum_i \log X_i-n\log \tau}=\frac{n}{\log \prod_i X_i-\log \tau^n}=\frac{n}{\log\frac{\prod_i X_i}{ \tau^n}}$$

Just note that $X_i$ are rv's, so capital letter is required. They become $x_i$ after seeing their realizations

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Now, if you want, you can now calculate also $\hat{\tau}$ supposing it is not known

Answer 2

We can also perform this calculation by considering the transformed random variable $Y_i = \log (X_i/\tau)$. Then the density of $Y$ becomes $$f_Y(y) = f_X(\tau e^y) \tau e^y = \theta \tau^\theta (\tau e^y)^{-(\theta+1)} \tau e^y = \theta e^{-\theta y}, \quad y \ge 0.$$ This of course is an exponential distribution with rate $\theta$. Then the MLE for $\theta$ is easily found: $$\mathcal L(\theta \mid \tau, \boldsymbol y) \propto \theta^n e^{-\theta n \bar y} \mathbb 1(y_{(1)} \ge 0),$$ hence $$\ell(\theta \mid \tau, \boldsymbol y) \propto \log \theta - \theta \bar y.$$ The derivative of the log-likelihood is $$\frac{\partial \ell}{\partial \theta} = \frac{1}{\theta} - \bar y,$$ with critical point (and maximum) $$\hat \theta \mid \tau = 1/\bar y.$$ Then by the invariance principle, $$\hat \theta \mid \tau = \frac{n}{\sum_{i=1}^n \log (x_i/\tau)}.$$ This assumes $\tau$ is known.

We could also do the MLE calculation directly on $X$, namely $$\mathcal L(\theta \mid \tau, \boldsymbol x) \propto \theta^n \tau^{n \theta} \left(\prod_{i=1}^n x_i \right)^{-\theta} \mathbb 1(x_{(1)} \ge \tau).$$ Note we discarded a constant factor of $\prod x_i$. Then $$\ell(\theta \mid \tau, \boldsymbol x) \propto \log \theta + \theta \log \tau - \frac{\theta}{n} \log \prod_{i=1}^n x_i.$$ Then $$\frac{\partial \ell}{\partial \theta} = \frac{1}{\theta} + \log \tau - \frac{1}{n} \sum_{i=1}^n \log x_i = \frac{1}{\theta} - \frac{1}{n} \sum_{i=1}^n \log (x_i/\tau)$$ and the critical point is again the same as above.

The density you call "weird" is known as a Pareto distribution and is quite common. We can see it arises naturally from a suitable transformation of an exponential distribution.