Aproxime la región crítica de manera que el tamaño de la prueba tiende a$\alpha$

Question

Considerar esta pregunta,

Supongamos $X_1, X_2, . . . , X_n$ es una muestra aleatoria de una distribución exponencial con una media de $\lambda$. Suponga que los datos observados se encuentra disponible en $[X_1], . . . , [X_n]$, en lugar de $X_1, . . . , X_n$, donde $[x]$ denota el mayor entero menor o igual a $x$. Considere la posibilidad de una prueba para $H_0 : \lambda = 1$ vs $H_1 : \lambda > 1$ que rechaza $H_0$ cuando $\displaystyle\sum_{i=1}^n [X_i ] > c_n$. Dado $\alpha \in (0, 1)$, obtener los valores de $c_n$ tal que el tamaño de la prueba converge a $\alpha$ como $n \to \infty$.

He sido capaz de averiguar que $[X_i]$ le siga distribución geométrica con parámetro de $(1 - e^{-\frac{1}{\lambda}})$ y tomará los valores de $0,1,2,...$. Como resultado, $\displaystyle\sum_{i=1}^n [X_i ]$ siendo la suma de $n$ geométrica de las variables aleatorias, siga con el Negativo de la Distribución binomial. Ahora el punto que yo estoy atascado en el cálculo del $c_n$. He intentado resolver mediante diversos enfoques (CLT por ejemplo), pero yo no estaba convencido por mis respuestas.

¿Cómo puedo resolver este problema?

Edit: estoy incluyendo aquí lo que yo hice con CLT

$\displaystyle \lim_{n\to\infty}P\left( \sum_{i=1}^n [X_i ] > c_n | H_0\right) \to \alpha$

$\displaystyle \lim_{n\to\infty}P\left( \frac{\sum_{i=1}^n [X_i ]}{n} > \frac{c_n}{n} | H_0\right) \to \alpha$

$\displaystyle \lim_{n\to\infty}P\left( \frac{\sum_{i=1}^n [X_i ]}{n} - \frac{e^{-1}}{1-e^{-1}}> \frac{c_n}{n} - \frac{e^{-1}}{1-e^{-1}}\right) \to \alpha$

$\displaystyle \lim_{n\to\infty}P\left( \frac{\frac{\sum_{i=1}^n [X_i ]}{n} - \frac{e^{-1}}{1-e^{-1}}}{\left(\frac{e^{-1}}{n(1-e^{-1})^2}\right)^\frac{1}{2}}> \frac{\frac{c_n}{n} - \frac{e^{-1}}{1-e^{-1}}}{\left(\frac{e^{-1}}{n(1-e^{-1})^2}\right)^\frac{1}{2}}\right) \to \alpha$

El uso de CLT,

$1 - \Phi\left( \dfrac{\frac{c_n}{n} - \frac{e^{-1}}{1-e^{-1}}}{\left(\frac{e^{-1}}{n(1-e^{-1})^2}\right)^\frac{1}{2}} \right) = \alpha$

$\Phi \left( \dfrac{c_n(1 - e^{-1}) - ne^{-1}}{\sqrt ne^{\frac{-1}{2}}}\right) = 1 - \alpha$

Deje $Z_\alpha$ ser tal que $\Phi(Z_\alpha) = 1 - \alpha$, luego

$\dfrac{c_n(1 - e^{-1}) - ne^{-1}}{\sqrt ne^{\frac{-1}{2}}} = Z_\alpha$, y, a continuación, la ecuación se resuelve para $c_n$.

Answer 1

Desde $X_i \sim \text{Exp}(\lambda)$ ha $\lfloor X_i \rfloor \sim \text{Geom}(1-e^{-\lambda})$, por lo que:

$$C_n = \sum_{i=1}^n \lfloor X_i \rfloor \sim \text{NegBin}(n, e^{-\lambda}).$$

(You have already established this part in your question.) All that you require in your question is an asymptotic size of $\alfa$, so this gives you a broad range of allowable sequences for the cut-off values $\{ c_n | n \in \mathbb{N} \}$. One way to proceed is to obtain an "optimal" test by setting the cut-off to the largest value that yields a size below the asymptotic level. Another is to use the CLT to obtain a form that gives the asymptotic result without worrying about its performance over smaller samples.

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Deriving the "optimal" cut-off level: To obtain a size level that is as close as possible to (but no greater than) the asymptotic level $\alpha$, we set the value $c_n$ to be the smallest value satisfying:

$$\begin{equation} \begin{aligned} \alpha \geqslant \mathbb{P} ( C_n > c_n | \lambda = 1 ) = 1-\sum_{r=0}^{c_n} \text{NegBin}(r|n,1/e). \end{aligned} \end{equation}$$

That is, we take:

$$c_n \equiv \min \Bigg\{ c =0,1,2,... \Bigg| 1-\sum_{r=0}^{c} \text{NegBin}(r|n,1/e) \leqslant \alpha \Bigg\}.$$

We will program this function below in R and look at the resulting size of the test as a function of $n$. Before we do this, we will also consider an alternative cut-off level using the central limit theorem.

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Deriving the cut-off level via the CLT: If we just want an approximation with the required asymptotic size then we can use the central limit theorem to obtain the approximation:

$$C_n \sim \frac{n}{e^\lambda-1} \cdot \text{N} \Big( 1, \frac{e^\lambda}{n} \Big).$$

Applying this approximation gives:

$$\begin{equation} \begin{aligned} \alpha \geqslant \mathbb{P} ( C_n > c_n | \lambda = 1 ) \approx 1 - \Phi \Bigg( \frac{(e-1) c_n/n - 1}{\sqrt{e/n}} \Bigg). \end{aligned} \end{equation}$$

Taking $c_n$ to be the smallest (real) value satisfying this approximate inequality, we have:

$$c_n = \frac{n}{e-1} \Bigg[ 1 + \sqrt{\frac{e}{n}} \cdot \Phi^{-1}(1-\alpha) \Bigg] = \frac{n}{e-1} \Bigg[ 1 + \mathcal{O}(n^{-1/2}) \Bigg].$$

This gives an explicit function for $c_n$ which has the specified asymptotic size.

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Programming and computation: We can program the "optimal" cut-off value and the approximation cut-off value as a functions of $n$ using the following R code.

library(stats);

#Function to implement test
TEST <- function(n, alpha, lambda) {
            e <-  exp(1);
            A <-  1;
            C <- -1;
            while(A > alpha) { 
                C <- C+1;
                A <- 1-pnbinom(C, n, 1-e^(-lambda), lower.tail = TRUE); }
            CC <- (n/(e^lambda-1))*(1 + sqrt(e^lambda/n)*qnorm(1-alpha));
            list(C = C, CC = CC, Size = A); }

To see how close the approximation is to the "optimal" cut-off we will plot both functions, and also plot the size of the optimal test. We can see from these plots that the size of the test converges to the chosen significance level as $n \rightarrow \infty$.

library(ggplot2);

#Generate cut-off values and size for test
N      <- 1000;
lambda <- 1;
alpha  <- 0.05
GRAPH  <- data.frame(n = 1:N, C = rep(0,N), CC = rep(0,N), Size = rep(0,N));
for (n in 1:N) { OUTPUTS       <- TEST(n, alpha, lambda);
                 GRAPH