Exercise: Let $X_1,\ldots,X_n$ be a random sample from the distribution with density $$f(x\mid\theta) = \dfrac{2x}{\theta^2}\mathbb{1}_{(0,\theta)}(x)$$ w.r.t. the Lebesgue measure. Derive an unbiased estimator for $\theta$. Does this estimator respect the likelihood principle?

I know that:

Def: (Likelihood principle (LP)).The information brought by an observation $x$ about $\theta$ is entirely contained in the likelihood function $L(\theta;x)$. Moreover, if $x$ and $x'$ are two observations depending on the same parameter (possibly in different experiments), such that there exists a constant $c$ satisfying $L(\theta;x) =cL'(\theta;x_0)$ for every $\theta$, they bring the same information about $\theta$ and must lead to identical inferences.

Question: Suppose I derived the unbiased estimator $\hat{\theta} = \dfrac{3}{2}X_1$; does this estimator satisfy the likelihood principle? I know that p-values do not respect the LP, because you reach different conclusions when using different p-values, but I'm not sure how an estimator would or would not respect the LP.

Thanks in advance!

  • 1
    $\begingroup$ Your estimator will not obey the likelihood principle if for example $X_1 \lt \frac23 X_2$, as in that case the likelihood of that $\hat{\theta}$ will be zero when there are other estimators with positive likelihood $\endgroup$ – Henry Jan 19 '18 at 8:52
  • $\begingroup$ Thank you for your comments guys. Unfortunately I don't really understand any better in what way an estimator can or cannot respect the LP. Could you elaborate a bit more? $\endgroup$ – titusAdam Jan 22 '18 at 10:51
  • $\begingroup$ @BruceET : Your comment does not appear to bear upon the question that was actually asked. Also, note this difference: $$ \begin{align} & \theta\mathsf{BETA}(2,1) & & \text{coded as \theta\mathsf{BETA}(2,1)} \\ & \theta\operatorname{\mathsf{BETA}}(2,1) & & \text{coded as \theta\operatorname{\mathsf{BETA}}(2,1)} \end{align} $$ Using \operatorname{} does not simply add some space; rather the spacing depends on the context. $\qquad$ $\endgroup$ – Michael Hardy Sep 13 '18 at 23:48
  • $\begingroup$ Sorry for typographical difficulties. Deleting comment. $\endgroup$ – BruceET Sep 14 '18 at 0:19

$$ L(\theta \mid X=x) = \begin{cases} \dfrac{2x}{\theta^2} & \text{for } \theta\in [x,+\infty), \\[8pt] \,\,0 & \text{for } \theta\in(0,x). \end{cases} $$ The MLE is therefore $x$ itself, since $L(\theta)$ increases as $\theta$ decreases UNTIL $\theta$ gets as small as $x.$ That is clearly a biased estimator.

Your unbiased estimator can be calculated if the MLE is known, and the MLE can be calculated if your estimator is known. Therefore, no information except what is in the likelihood function goes into your estimator.

Merely being unbiased is not enough to guarantee violation of the likelihood principle. But an insistence on always being unbiased will result in some estimators violating the likelihood principle.


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