# Does this estimator respect the likelihood principle?

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.

• 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 – Henry Jan 19 '18 at 8:52
• @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. $\qquad$ – Michael Hardy Sep 13 '18 at 23:48
$$L(\theta \mid X=x) = \begin{cases} \dfrac{2x}{\theta^2} & \text{for } \theta\in [x,+\infty), \\[8pt] \,\,0 & \text{for } \theta\in(0,x). \end{cases}$$ 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.