Suppose we have a random sample from the distribution $f(x|\lambda) = \lambda x^{-2}$ with $x > \lambda$. I want to find a UMP level $\alpha$ test for the hypothesis test $H_0: \lambda = \lambda_0$ and $H_1: \lambda = \lambda_1$, where $\lambda_1 > \lambda_0$.

I believe $\prod_{i=1}^n x_i$ is a sufficient statistic for $\lambda$, but I don't know what to do from here. What is the general strategy?


1 Answer 1


This distribution is a Pareto distribution with shape parameter equal to $1$.

While we are stepping through the derivation of the uniformly most powerful test, we'll make use of two facts about the Pareto distribution.

First, as it happens, the minimal sufficient statistics for the Pareto distribution with unknown lower bound $\lambda$ and unknown shape parameter $a$ are the minimum value of $x$ ($x_{[1]}$) and $\sum\log x$ (see Estimation of the parameters of the Pareto distribution.) If $a$ is known, as it is in this case, then $x_{[1]}$ is sufficient for $\lambda$.

As a result, we will be interested in the distribution of $x_{[1]}$. Our second fact is that the distribution of the minimum value from a collection of Pareto distributions with lower bound $\lambda$ and shape parameters $a_i, i=1, \dots, N$ is Pareto with shape parameter $a_0 = \sum_{i=1}^N a_i$ and lower bound $\lambda$. For this problem this equates to a $\text{Pareto}(n,\lambda)$ distribution.

On to the question! The Neyman-Pearson lemma tells us that the UMP test for a simple hypothesis and alternative, as in the problem statement, is a likelihood ratio test. Since $x_{[1]}$ is sufficient for the parameter of interest and the other parameter is known, the test can be constructed using the distribution for $x_{[1]}$ under the null and alternative hypotheses. Writing out the likelihood ratio gives us:

$$\Lambda = {\mathcal{L}(\lambda_0;x_{[1]}) \over \mathcal{L}(\lambda_1;x_{[1]}) } = {\lambda_0x_{[1]}^{-n-1}1(x_{[1]} \geq \lambda_0) \over \lambda_1x_{[1]}^{-n-1}1(x_{[1]} \geq \lambda_1)} = {\lambda_01(x_{[1]} \geq \lambda_0) \over \lambda_11(x_{[1]} \geq \lambda_1)}$$

where $1(x_{[1]} \geq \lambda_0)$ is the indicator function that takes on the value $1$ if the condition is true, $0$ otherwise.

Since $\lambda_1 > \lambda_0$, the likelihood ratio takes on the value $\lambda_0/\lambda_1$ when $x_{[1]} \geq \lambda_1$ and is undefined when $x_{[1]} < \lambda_1$. In the latter case, we know that the alternative hypothesis is false, so we would not want to reject the null hypothesis.

Now for constructing an $\alpha$-level test. We consider two cases:

  1. $P(x_{[1]} < \lambda_1|H_0) \geq 1-\alpha$,
  2. $P(x_{[1]} < \lambda_1|H_0) < 1-\alpha$

In the special case (as here) of $a = 1$, $P(x_{[1]} < \lambda_1 | H_0) = 1 - \lambda_0/\lambda_1$, and a small amount of algebra reveals that these conditions simplify to:

  1. ${\lambda_0 \over \lambda_1} \leq \alpha$,
  2. ${\lambda_0 \over \lambda_1} > \alpha$

In the first case, the probability of incorrectly rejecting $H_0$ is $\leq \alpha$ even if we reject 100% of the time that $x_{[1]} \geq \lambda_1$. Therefore, the closest we can get to an $\alpha$ level test is to reject $H_0$ all the time. The actual level of this test is $\lambda_0 / \lambda_1$.

In the second case, we need to construct a rejection region in $[\lambda_1, \infty)$ with the appropriate probability. Usually we would take the region(s) with the lowest likelihood ratios, writing imprecisely, because this gives us the best type II error, but in this case the likelihood ratio is constant over the entire region, so no guidance is to be found there. On the other hand, this fact means that any region with the appropriate probability under $H_0$ is as good as any other, so we may as well pick a region of the form $[\lambda_1, \tau]$.

We want our rejection region to satisfy:

$${P(\lambda_1 \leq x_{[1]} \leq \tau|H_0) \over P(\lambda_1 \leq x_{[1]} \leq \tau|H_0) + 1 - \lambda_0/\lambda_1} = \alpha$$

Some algebra shows us that:

$$P(\lambda_1 \leq x_{[1]} \leq \tau|H_0) = {\lambda_0 \over \lambda_1} - {\lambda_0 \over \tau}$$

which, when combined with the above and a little simplification of the denominator, leads to:

$${{\lambda_0 \over \lambda_1} - {\lambda_0 \over \tau} \over 1-{\lambda_0 \over\tau}} = \alpha$$

Multiplying both numerator and denominator on the l.h.s. by $\tau$ gives us:

$${{\lambda_0 \over \lambda_1}\tau - \lambda_0 \over \tau - \lambda_0} = \alpha$$

and some further rearrangement gives us:

$$\tau_{\alpha} = {(1-\alpha)\lambda_0 \over {\lambda_0 \over \lambda_1} - \alpha}$$

The sign of the denominator need not worry us, as we know from case 2 above that $\lambda_0/\lambda_1 > \alpha$.

We can show that $\tau > \lambda_1$ easily enough by multiplying the top and bottom of the r.h.s. by $\lambda_1$ and observing that:

$$\tau > \lambda_1 \text{ iff } {(1-\alpha)\lambda_0\lambda_1 \over \lambda_0 - \alpha\lambda_1} > \lambda_1$$

We check whether the ratio:

$${(1-\alpha)\lambda_0 \over \lambda_0 - \alpha\lambda_1} = {\lambda_0-\alpha\lambda_0 \over \lambda_0 - \alpha\lambda_1} > 1$$

which, since $\lambda_1 > \lambda_0$, is clearly true.

So, in the case where $\lambda_0/\lambda_1 > \alpha$, the $\alpha$-level UMP is to reject the null hypothesis when $x_{[1]} \in [\lambda_1, \tau_{\alpha}]$, with $\tau_{\alpha}$ defined as above.


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