Suppose $X_1, . . . ,X_n$ are a random sample from a population whose density function is $f(x|β) = \frac{β}{x^{β+1}}$ , $1 < x < ∞$, $β > 0$.

Question

For the distribution, $f(x|β) = \frac{β}{x^{β+1}}$ , $1 < x < ∞$, $β > 0$.

(a) Find the uniformly most powerful test of $H_0 : β = 1$ against $H_1 : β < 1$.

I have found it as $\prod{X_i} > c$ via likelihood ratio test.

(b) Conduct the test, giving a $p$-value and stating your conclusions clearly, based on the following 20 observed values:
1.23 2.74 4.99 5.11 5.55 6.55 138.80 1.90 4.74 2.53
2.41 1.21 26.55 6.81 1.17 1.39 1.08 4.87 2.91 2.57

I was struggling to find the distribution of $\prod X_i$ to calculate $p$-value, are there any good ideas?

Answer 1

Thanks for your hint @henry , I may have a result post as below.

For question (a)

Consider a simple statistical hypotheses, $H_0:\beta = 1$ against $H_1:\beta=\beta_1<1$. Then the likelihood ratio is as
$$ \frac{L(1)}{L(\beta_1)}=\frac{1}{\prod{x_i^2}}\frac{\prod{x_i^{\beta_1+1}}}{\beta_1^{n}}=\frac{\prod{x_i^{\beta_1-1}}}{\beta_1^{n}} $$ Let the ratio $$ \begin{aligned} \frac{L(1)}{L(\beta_1)} &\leq k \\ \frac{\prod{x_i^{\beta_1-1}}}{\beta_1^{n}} &\leq k \\ (\beta_1-1)\sum\ln{x_i} &\leq \ln{k}+n\ln{\beta_1}\\ \sum\ln{x_i} &\geq c \end{aligned} $$ Hence we get a uniformly most powerful region for testing.

For question (b)

Let us consider the distribution of $Y=\ln{X}$ first. Using the transformation equation of pdf we have $$ \begin{aligned} f_Y(y) &= f_X(g^{-1}(y))|\frac{d}{dy}g^{-1}(y)| \\ &= \frac{\beta}{e^{\beta y+y}}e^y \\ &= \beta e^{-\beta y} \end{aligned} $$ Which is a exponential distribution $Exp(\beta)$ or a gamma distribution $\Gamma (1,\frac{1}{\beta})$, so we know the distrubution of sum of y $$\sum{y_i}=\sum{\ln{x_i}} \sim \Gamma (n,\frac{1}{\beta})$$ According the observed data, we can finally calculate a p-value as follow: $$ \begin{aligned} p &= P(\sum{\ln{X_i}} \geq \sum{\ln{x_i|H_0}}) \\\\ &= P(\sum{\ln{X_i}} \geq 26.5321|\beta = 1) \\\\ &= 1-F_{\Gamma}(x=26.5321,\alpha=20,\beta = 1) \\\\ &= 0.08088 \nonumber \end{aligned} $$ where $F_{\Gamma}$ is the cdf of gamma distribution.
So we can reject $H_0$ at confidence level $\alpha = 0.1$, but accept it if $\alpha = 0.05 $