Theoretical Solution
Theorem
Let \(\Theta \subset \mathbb{R}\) and suppose \(P_\theta\) has a density \(f_\theta\) with respect to some fixed measure \(\mu\), belonging to a one-dimensional exponential family of the form:
\[ f_\theta(x) = h(x)\exp\{C(\theta)T(x) - B(\theta)\} \]
Define the sufficient statistic for \(\theta\) as: \[ S(\mathbf{X}) = \sum_{i=1}^n T(X_i) \]
Then:
For testing \(H_0: \theta \in [\theta_a, \theta_b]\) versus \(H_1: \theta \notin [\theta_a, \theta_b]\), the test has the form: \[ \varphi(\mathbf{X}) = \begin{cases} 1, & \text{if } S(\mathbf{X}) \notin [c_a, c_b] \\ 0, & \text{if } S(\mathbf{X}) \in [c_a, c_b] \end{cases} \] where constants \(c_a, c_b\) are chosen so that: \[ \mathbb{E}_{\theta_a}[\varphi(\mathbf{X})] = \mathbb{E}_{\theta_b}[\varphi(\mathbf{X})] = \alpha \] This is the uniformly most powerful unbiased test.
For testing \(H_0: \theta = \theta_0\) versus \(H_1: \theta \neq \theta_0\), the test has the same form: \[ \varphi(\mathbf{X}) = \begin{cases} 1, & \text{if } S(\mathbf{X}) \notin [c_a, c_b] \\ 0, & \text{if } S(\mathbf{X}) \in [c_a, c_b] \end{cases} \] with constants \(c_a, c_b\) chosen so that: \[ \mathbb{E}_{\theta_0}[\varphi(\mathbf{X})] = \alpha, \quad \mathbb{E}_{\theta_0}[\varphi(\mathbf{X}) S(\mathbf{X})] = \alpha \cdot \mathbb{E}_{\theta_0}[S(\mathbf{X})] \]
This test is also uniformly most powerful among all unbiased tests.
Our Test
In our case, the exponential distribution belongs to the exponential
family. The sufficient statistic is:
\(T(X) = \sum_{i=1}^{n}X_i\)
We will test the hypothesis \(H_0: \lambda = 4\) vs. \(H_1: \lambda \neq 4\)
For simplicity, let \(k =
1/\lambda\).
Then we have: \(X_1, ..., X_n\) i.i.d.
from \(\text{Exp}(k)\), and the
expected value is \(1/k =
\lambda\).
The test has the form: \[ \varphi(\mathbf{X}) = \begin{cases} 1, & \text{if } T(\mathbf{X}) \notin [a, b] \\ 0, & \text{if } T(\mathbf{X}) \in [a, b] \end{cases} \] where constants \(a, b\) are chosen such that: \[ \mathbb{E}_{\theta_0}[\varphi(\mathbf{X})] = \alpha, \quad \mathbb{E}_{\theta_0}[\varphi(\mathbf{X}) T(\mathbf{X})] = \alpha \cdot \mathbb{E}_{\theta_0}[T(\mathbf{X})] \]
Distribution of the Test Statistic under the Null Hypothesis
\(X_1, ..., X_n\) is a random sample
from \(\text{Exp}(k)\).
Therefore, \(T(X) = \sum_{i=1}^{n}X_i \sim
\text{Gamma}(n, k)\)
Finding the Constants \(a, b\), with \(a < b\)
First Condition
\[ \mathbb{E}_{\theta_0}[\varphi(\mathbf{X})] = \alpha \]
\[ 1 \cdot P_{\theta_0}(T(X) \notin (a,b)) = \alpha \]
\[ P_{\theta_0}(T(X) \in (a,b)) = 1 - \alpha \]
\[ P_{\theta_0}(a < T(X) < b) = 1 - \alpha \]
Under the null, \(T(X) \sim \text{Gamma}(n, k)\), so:
\[ F_{\text{Gamma}(n, k)}(b) - F_{\text{Gamma}(n, k)}(a) = 1 - \alpha \]
Second Condition
\[ \mathbb{E}_{\theta_0}[\varphi(\mathbf{X}) T(\mathbf{X})] = \alpha \cdot \mathbb{E}_{\theta_0}[T(\mathbf{X})] \]
Assume \(Y \sim \text{Gamma}(n, k)\). Then:
\[ f_Y(t) = \frac{k^n}{\Gamma(n)} t^{n-1} e^{-kt} \cdot \mathbb{1}_{(0, \infty)}(t) \]
\[ \mathbb{E}[Y] = \frac{n}{k} \]
Right-hand side:
\[ \alpha \cdot \mathbb{E}_{\theta_0}[T(\mathbf{X})] = \alpha \cdot \frac{n}{k} \]
Substitute into the second expectation:
\[ \mathbb{E}_{\theta_0}[\varphi(\mathbf{X}) T(\mathbf{X})] = \alpha \cdot \frac{n}{k} \]
Written as an integral (omitting steps similar to the first condition):
\[ \int_{a}^{b} t \cdot t^{n-1} e^{-kt} \, dt = (1 - \alpha) \cdot \frac{n}{k} \cdot \frac{\Gamma(n)}{k^n} \]
\[ \int_{a}^{b} t^n e^{-kt} \, dt = (1 - \alpha) \cdot \frac{\Gamma(n+1)}{k^{n+1}} \]
\[ \int_{a}^{b} \frac{k^{n+1}}{\Gamma(n+1)} t^n e^{-kt} \, dt = 1 - \alpha \]
\[ F_{\text{Gamma}(n+1, k)}(b) - F_{\text{Gamma}(n+1, k)}(a) = 1 - \alpha \]
Final System of Equations
We solve the following system numerically to find the constants \(a, b\):
\[ \begin{cases} F_{\text{Gamma}(n, k)}(b) - F_{\text{Gamma}(n, k)}(a) = 1 - \alpha \\ F_{\text{Gamma}(n+1, k)}(b) - F_{\text{Gamma}(n+1, k)}(a) = 1 - \alpha \end{cases} \]
One-Sample Test
Parameters
alpha <- 0.05
n <- 20
# Null hypothesis parameter
lambda <- 4
k <- 1 / lambda
# Parameter for the generated sample
lambda2 <- 2
k2 <- 1 / lambda2Checking the distribution generators in R
## [1] 0.2123532## [1] 3.992103Generating a sample and computing the test statistic
## [1] 35.21126Finding the constants \(a,b\)
uklad <- function(rownania){
a <- rownania[1]
b <- rownania[2]
eq1 <- pgamma(b, n+1, k) - pgamma(a, n+1, k) - (1 - alpha)
eq2 <- pgamma(b, n, k) - pgamma(a, n, k) - (1 - alpha)
c(eq1, eq2)
}
# Initial guess near the mean
m <- n / k
solution <- nleqslv(c(m - 10, m + 10), uklad)
solution$x## [1] 49.75731 120.54966Does the test statistic fall outside the critical region?
## [1] 35.21126## [1] TRUEEmpirical Power for Given Parameters
Simulation Setup
MC = 1000000
alpha <- 0.05
lamdba <- 4
k <- 1 / lamdba
n <- 20
# Lambda values for the power curve
lamdamoc <- seq(2, 6, by = 0.05)Finding constants \(a,b\) again
uklad <- function(rownania){
a <- rownania[1]
b <- rownania[2]
eq1 <- pgamma(b, n+1, k) - pgamma(a, n+1, k) - (1 - alpha)
eq2 <- pgamma(b, n, k) - pgamma(a, n, k) - (1 - alpha)
c(eq1, eq2)
}
m <- n / k
solution <- nleqslv(c(m - 10, m + 10), uklad)
solution$x## [1] 49.75731 120.54966Running the Monte Carlo Simulation for each \(\lambda\)
empmoc <- c()
for(i in 1:length(lamdamoc)){
TMC <- c()
for(j in 1:MC){
X <- rexp(n, 1 / lamdamoc[i])
TMC[j] <- sum(X)
}
empmoc[i] <- sum(TMC < a | TMC > b) / MC
}Plotting the Empirical Power as a Function of \(\lambda\)
df <- data.frame(lamdamoc, empmoc)
ggplot(df, aes(x = lamdamoc, y = empmoc)) +
geom_hline(yintercept = 0.05, color = "darkred", linewidth = 1) +
geom_point(color = "darkblue") +
theme_minimal() +
labs(x = "Lambda", y = "Empirical Power", title = "Power Curve") +
scale_y_continuous(limits = c(0, max(df$empmoc, na.rm = TRUE)))