Solution: Deriving the Uniformly Most Powerful (UMP) Test

The problem asks us to find a uniformly most powerful (UMP) test of size $\alpha$ for testing the null hypothesis $H_0: \theta = 1$ against the alternative hypothesis $H_1 : \theta > 1$. The data consists of a random sample $X_1, X_2, \ldots, X_n$ drawn from a continuous distribution with the probability density function (PDF):

$ f(x|\theta) = \begin{cases} \frac{2\theta^2}{x^{\theta+1}}, & \text{if } x>2, \\ 0, & \text{otherwise,} \end{cases} $

where $\theta > 0$ is an unknown parameter.

Understanding Uniformly Most Powerful (UMP) Tests

A UMP test is a test that is most powerful for all possible parameter values in the alternative hypothesis. For a one-sided hypothesis test like this one, a UMP test often exists if the distribution belongs to a family with a monotone likelihood ratio in a sufficient statistic. The Neyman-Pearson lemma and its extensions guide us in constructing such tests.

Constructing the Likelihood Ratio

First, we write down the likelihood function for the sample $x_1, \ldots, x_n$:

$ L(\theta) = \prod_{i=1}^n f(x_i|\theta) = \prod_{i=1}^n \frac{2\theta^2}{x_i^{\theta+1}} = (2\theta^2)^n \prod_{i=1}^n x_i^{-(\theta+1)} $

To simplify analysis, we consider the log-likelihood function:

$ \ln L(\theta) = \ln \left( (2\theta^2)^n \prod_{i=1}^n x_i^{-(\theta+1)} \right) $

$ \ln L(\theta) = n \ln(2\theta^2) - (\theta+1) \sum_{i=1}^n \ln X_i $

$ \ln L(\theta) = n (2\ln\theta + \ln 2) - (\theta+1) \sum_{i=1}^n \ln X_i $

Monotone Likelihood Ratio Property

The distribution belongs to the exponential family. We can write the log-likelihood in the form $\eta(\theta) T(x) + A(\theta) + B(x)$, where $T(x)$ is a statistic. Let's rearrange the log-likelihood:

$ \ln L(\theta) = -(\theta+1) \left( \sum_{i=1}^n \ln X_i \right) + 2n \ln \theta + n \ln 2 $

Here, we can identify:

• Statistic: $T = \sum_{i=1}^n \ln X_i$
• Parameter function: $\eta(\theta) = -(\theta+1)$
• Other terms: $A(\theta) = 2n \ln \theta$, $B(x) = n \ln 2$ (constant with respect to $\theta$)

To check for a monotone likelihood ratio, we examine the derivative of $\eta(\theta)$ with respect to $\theta$:

$ \frac{d\eta(\theta)}{d\theta} = \frac{d}{d\theta} (-(\theta+1)) = -1 $

Since $\frac{d\eta(\theta)}{d\theta} = -1 < 0$, the function $\eta(\theta)$ is strictly decreasing in $\theta$. For a family with a strictly decreasing $\eta(\theta)$, the UMP test for $H_0: \theta = \theta_0$ against $H_1: \theta > \theta_0$ rejects $H_0$ if the statistic $T$ takes *small* values. In our case, $H_0: \theta = 1$ and $H_1: \theta > 1$, so we reject $H_0$ if $\sum_{i=1}^n \ln X_i$ is sufficiently small.

Distribution of the Test Statistic under $H_0$

We need to find the distribution of the test statistic $T = \sum_{i=1}^n \ln X_i$ when $H_0$ is true ($\theta=1$). Under $H_0$, the PDF is $f(x|1) = \frac{2(1)^2}{x^{1+1}} = \frac{2}{x^2}$ for $x > 2$.

Let's transform the random variable $X_i$. Consider $Z_i = \ln X_i$. Since $X_i > 2$, we have $Z_i > \ln 2$. The cumulative distribution function (CDF) of $Z_i$ is:

$ F_{Z_i}(z) = P(Z_i \le z) = P(\ln X_i \le z) = P(X_i \le e^z) $

For $z > \ln 2$ (which implies $e^z > 2$):

$ F_{Z_i}(z) = \int_2^{e^z} \frac{2}{t^2} dt = 2 \left[-\frac{1}{t}\right]_2^{e^z} = 2 \left(-\frac{1}{e^z} - \left(-\frac{1}{2}\right)\right) = 1 - \frac{2}{e^z} $

Now, let's make another transformation to simplify the distribution. Let $V_i = Z_i - \ln 2$. Then $V_i > 0$. The CDF of $V_i$ is:

$ F_{V_i}(v) = P(V_i \le v) = P(Z_i - \ln 2 \le v) = P(Z_i \le v + \ln 2) $

For $v > 0$:

$ F_{V_i}(v) = 1 - \frac{2}{e^{v+\ln 2}} = 1 - \frac{2}{e^v \cdot e^{\ln 2}} = 1 - \frac{2}{e^v \cdot 2} = 1 - e^{-v} $

This is the CDF of an Exponential distribution with rate parameter $\lambda=1$, i.e., $V_i \sim \text{Exp}(1)$.

The test statistic is $T = \sum_{i=1}^n \ln X_i = \sum_{i=1}^n Z_i$. We can express $T$ in terms of $V_i$:

$ T = \sum_{i=1}^n (V_i + \ln 2) = \sum_{i=1}^n V_i + n \ln 2 $

Let $S = \sum_{i=1}^n V_i$. Since $V_i$ are i.i.d. Exponential(1) random variables, their sum $S$ follows a Gamma distribution with shape parameter $n$ and scale parameter $1$. That is, $S \sim \text{Gamma}(n, 1)$.

A known property is that if $S \sim \text{Gamma}(n, 1)$, then $2S \sim \chi^2_{2n}$ (a chi-squared distribution with $2n$ degrees of freedom).

Therefore, under $H_0$, the statistic $T$ has the distribution:

$ T = S + n \ln 2 = \frac{1}{2} (2S) + n \ln 2 $

$ T \sim \frac{1}{2} \chi^2_{2n} + n \ln 2 $

Determining the Critical Region

We established that for $H_1: \theta > 1$, we should reject $H_0$ if the test statistic $T = \sum_{i=1}^n \ln X_i$ is small. The rejection region is of the form $T \le c$, where $c$ is the critical value determined by the size $\alpha$ of the test. This means:

$ P(T \le c | \theta = 1) = \alpha $

Substitute the distribution of $T$ under $H_0$:

$ P\left(\frac{1}{2} \chi^2_{2n} + n \ln 2 \le c \right) = \alpha $

Rearranging the inequality inside the probability:

$ P\left(\frac{1}{2} \chi^2_{2n} \le c - n \ln 2 \right) = \alpha $

$ P\left(\chi^2_{2n} \le 2(c - n \ln 2) \right) = \alpha $

Now, we need to interpret the quantile notation used in the question. The question states: "$P(Y>\chi^2_{m,\beta}) = \beta$, where $Y \sim \chi^2_m$." This means $\chi^2_{m,\beta}$ represents the *upper* $\beta$-quantile of the chi-squared distribution with $m$ degrees of freedom. This is also known as the $(1-\beta)$-th quantile in standard notation.

The equation $P\left(\chi^2_{2n} \le 2(c - n \ln 2) \right) = \alpha$ means that $2(c - n \ln 2)$ must be the *lower* $\alpha$-quantile of the $\chi^2_{2n}$ distribution.

The lower $\alpha$-quantile is the value $k'$ such that $P(\chi^2_{2n} \le k') = \alpha$. This is equivalent to $P(\chi^2_{2n} > k') = 1 - \alpha$. Using the notation from the question, this value is precisely $\chi^2_{2n, 1-\alpha}$.

Therefore, we set:

$ 2(c - n \ln 2) = \chi^2_{2n, 1-\alpha} \quad \text{(using the question's quantile notation)} $

Solving for $c$:

$ c - n \ln 2 = \frac{1}{2} \chi^2_{2n, 1-\alpha} $

$ c = \frac{1}{2} \chi^2_{2n, 1-\alpha} + n \ln 2 $

The rejection region for $H_0$ is $T \le c$. Thus, we reject $H_0$ if:

$ \sum_{i=1}^n \ln X_i \le \frac{1}{2} \chi^2_{2n, 1-\alpha} + n \ln 2 $

Conclusion

The uniformly most powerful test of size $\alpha$ rejects $H_0: \theta = 1$ against $H_1 : \theta > 1$ if $\sum_{i=1}^n \ln X_i \le \frac{1}{2} \chi^2_{2n, 1-\alpha} + n \ln 2$. This corresponds to option 2.