Suppose $X_1, \dots, X_n \overset{\text{iid}}{\sim}\dfrac{x}{\theta}\exp\left(-\dfrac{x^2}{2\theta}\right)\mathbf{1}_{(0, \infty)}(x)$, $\theta > 0$. At the end of the day, my goal is to calculate $$\max_{\theta > 16}L(\theta)$$ where $L$ is the likelihood function, i.e., $$L(\theta) = \dfrac{\prod_{i=1}^{n}x_i}{\theta^n}\exp\left(\dfrac{-1}{2\theta}\sum_{i=1}^{n}x_i^2\right)\text{.}$$ In order to do this, we have to find $\theta$ such that $L$ is maximized. Using the typical methods, I find that $$\hat{\theta}_n = \dfrac{\sum_{i=1}^{n}X_i^2}{2n}$$ is the maximum likelihood estimator of $\theta$, assuming $\theta > 0$.

However, when restricting $\theta$ to $\{\theta: \theta > 16\}$, the solution I have says that the maximum likelihood estimator is $$\hat{\theta}_n = \max\left(16, \dfrac{\sum_{i=1}^{n}X_i^2}{2n}\right)$$ is the MLE. Why is this? Intuitively, this doesn't make any sense to me because one would think that one wants $\hat\theta_n$ to be as small as possible, since increasing $\theta$ would decrease $L$, holding the $\{x_i\}$ constant. So I think $\max$ should be replaced with $\min$ in the equation for $\hat\theta_n$ above. However, yet at the same time, we are restricting $\theta > 16$.

Could someone please provide me with some insight for this?

  • $\begingroup$ @Vim Because 1) outside of the $\exp$ term, it's in the denominator of a fraction - by increasing $\hat\theta$, you decrease $L$ and 2) inside the $\exp$ term, by increasing $\hat\theta$, you... oh wait, that would increase $L$ because of the negative sign. So it's not as simple as I'm making it, just adding to my confusion even more... $\endgroup$ – Clarinetist May 9 '17 at 13:55
  • $\begingroup$ I don't think one would want $\theta$ as small as possible, note that the likelihood is not decreasing in $\theta$. Otherwise you couldn't have possibly found the ML estimator by using the "typical" method which I assume means differentiation. $\endgroup$ – Vim May 9 '17 at 13:57
  • $\begingroup$ @Vim Yeah, I'll have to look into this more, maybe perhaps by looking closer at the first partial of the loglikelihood. $\endgroup$ – Clarinetist May 9 '17 at 13:58

$L(\theta)$ goes up when $\theta$ goes from $0$ to the typical MLE point (the first $\hat\theta$ you get), reaches peak at $\hat\theta$ and goes down henceforth. This should give you the idea.

  • $\begingroup$ Let's consider the following analysis. Consider the formula $\hat\theta_n$ that I calculated using the typical differentiation method with $\theta > 0$. By setting the restriction $\theta > 16$, all we've essentially done is truncated $L$ so that any values of $\{L(\theta): \theta \leq 16\}$ are now all $0$. Suppose $\hat\theta_n > 16$. Then $\hat\theta_n$ remains the MLE because it's still the peak of the graph. Now suppose $\hat\theta_n < 16$. Then the likelihood is strictly decreasing, and thus $16$ is the MLE. Are my thoughts correct? $\endgroup$ – Clarinetist May 9 '17 at 14:07
  • $\begingroup$ @Clarinetist yeah that's exactly what I mean. $\endgroup$ – Vim May 9 '17 at 14:09
  • $\begingroup$ Thank you! Now if I can remember to think about this while I'm taking this qualifying exam. I appreciate your help! $\endgroup$ – Clarinetist May 9 '17 at 14:09
  • $\begingroup$ @Clarinetist you're very welcome. $\endgroup$ – Vim May 9 '17 at 14:10

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