Solution for Exercise 9.44 (a) in Statistical Inference (Casella & Berger)

This page discusses the problem of constructing a one-sided $(1-\alpha)$ UMA confidence interval for the parameter $\lambda$ of a Poisson distribution, based on the level $\alpha$ UMP test for hypotheses $H_0: \lambda=\lambda_0$ vs $H_1: \lambda>\lambda_0$ .

(中文)
本人是在台灣就讀數學系博士班的日本學生。在國立台灣大學擔任統計理論(高等統計理論)的助教時，第二學期某一次作業有一題是來自Casella and Berger撰寫的數理統計書《Statistical Inferece》中的習題：

Constructing a one-sided $(1-\alpha)$ UMA confidence interval for a Poisson parameter. 出自:《Statistical Inference》 (Casella & Berger) Exercise 9.44 (a):
Let $X_1, \cdots, X_n \stackrel{\rm iid}{\sim} {\rm Po}(\lambda)$ . Find a UMA $1-\alpha$ confidence interval based on inverting the UMP level $\alpha$ test of $H_0: \lambda=\lambda_0$ vs $H_1: \lambda > \lambda_0$ .

我在批改作業時發現，所有人的答案都跟我算出的答案不一樣。幾乎所有人算出來的結果是 $[0, \frac{1}{2n}\chi^2_{2(T+1),\alpha}]$ ，其中 $\chi^2_{2(T+1),\alpha}$ 為 $(1-\alpha)$ th quantile of 卡方分布 with $2(T+1)$ degrees of freedom and $T=\sum_{i=1}^n X_i$ .

然而我自己算出來的答案卻是 $[\frac{1}{2n}\chi^2_{2T,1-\alpha},\infty)$ 。大家寫的答案是 $[0, U(\bm{x})]$ 的形式，而我得到的答案則是 $[L(\bm{x}), \infty)$ 的形式。

為了初步的觀察，我們可以先把題目的分布改為 $N(\lambda,1)$ 。無論是常態分布還是泊松分布，當參數越大，我們所觀測的值也會越大，所以可以說是參數與樣本的大小關係是相同的。在這種情況下，假使今天把題目中的分布換成常態分布，我們得到的信賴區間的形式應該會一樣的。為了簡單起見，考慮 $n=1$ 的情況，此時拒絕域為 $R(\lambda_0) = \{x \mid x-\lambda_0 > \Phi^{-1}(1-\alpha)\}$ ，因此接受域為 $A(\lambda_0) = \{x \mid x-\lambda_0 \leq \Phi^{-1}(1-\alpha)\}$ 。我們得到信賴區間 $C(x) = \{\lambda_0 \mid x-\lambda_0 \leq \Phi^{-1}(1-\alpha)\} =[x-\Phi^{-1}(1-\alpha),\infty)$ 。由此可以預期，即使在泊松分布的情況下，我們仍會得到 $[L(\bm{x}), \infty)$ 的形式。

後來發現大家可能參考了以下網站(stackexchange)寫的東西:
https://stats.stackexchange.com/questions/342824/what-is-the-uma-1-alpha-confidence-interval-around-lambda-given-sample-iid-poiss

這個網頁的方法並沒有直接將接受域寫成信賴區間，然後中間出現了一個隨機變數 $U$ ，不過我們觀測到的樣本是 $X_1, \ldots, X_n$ ， $U$ 似乎域 $X_1, \ldots, X_n$ 沒有關係。

下面我將提供我的做法，如果有人寫作業時參考了我的答案，請明示出處。

P.S. 順帶一提，誰和誰互相抄作業，或者哪些人直接抄了網頁上的解答，其實都一目了然。同學之間互相討論或參考書籍、網頁是很正常的事，不過我們最後應該自己消化，用自己的語言來重新將自己的思路呈現出來，這樣才會對學習有幫助。

(English and Japanese)

I am a Japanese PhD student studying in Taiwan and majoring in statistics in the Department of Mathematics. When I was a TA for the statistical theory course at National Taiwan University, an exercise from Casella & Berger’s “Statistical Inference” (Exercise 9.44 (a)) was assigned as homework.

I noticed that my solution differed from those written by the students. They likely referred to a method found on Stack Exchange, but I believe that both the method and the result are incorrect. Therefore, I am providing my solution below. If you use my solution as a reference for your homework submission, please be sure to cite the source (this page).

台大の統計理論(元・高等統計推論)のTAをやっていた時に宿題にCasella & BergerのExercise 9.44 (a)が出題された。この問題はポアソン分布のパラメータに対して与えられた仮説に対する一様最強力検定を考えることで、最精密片側信頼区間(one-sided UMA confidence interval)を構成せよという問題であったが、みんなの解答が私の解答と異なっていたので、このページに私の解答を掲載することにします。もしこのページを参考にされた場合は出典としてこのページを明記してくださるようお願いします。

(Here is my solution)

By an easy extension of the Neyman-Pearson’s lemma, we can easily see that the following test is a level $\alpha$ UMP-test for the given hypotheses in the problem.

\varphi(\bm{x}) \stackrel{\rm def}{=} \mathbb{1}(\{\bm{x}\mid T(\bm{x}) > k(\lambda_0)\}),

where $T(\bm{x}) \stackrel{\rm def}{=} \sum_{i=1}^n x_i$ and $k(\lambda_0)$ is a function with respect to $\lambda_0$ defined by the equation below:

k(\lambda_0) \stackrel{\rm def}{=} {\rm inf}\{j \mid \sum_{t=j+1}^\infty e^{-n\lambda_0} \frac{(n\lambda_0)^t}{t!} \leq \alpha\}.

Therefore the acceptance region is given by $A(\lambda_0) = \{\bm{x} \mid T(\bm{x}) \leq k(\lambda_0)\}$ , thus we further obtain the corrsponding confidence set as $C(\bm{x}) = \{\lambda_0 \mid T(\bm{x}) \leq k(\lambda_0)\}$ .

Next, we simplify the expression of the above confidence set (interval). From the definition of $k(\lambda_0)$ , we have the following obervation:

$T(\bm{x}) \leq k(\lambda_0) \Leftrightarrow T(\bm{x})-1 \notin \{j \mid \sum_{t=j+1}^{\infty} e^{-n\lambda_0} \frac{(n\lambda_0)^t}{t!} \leq \alpha\} \Leftrightarrow \sum_{t=T(\bm{x})}^\infty e^{-n\lambda_0} \frac{(n\lambda_0)^t}{t!} > \alpha.$ .

It follows that the confidence set above can further be rewritten by:
$C(\bm{x}) = \{\lambda_0 \mid \sum_{t=T(\bm{x})}^\infty e^{-n\lambda_0} \frac{(n\lambda_0)^t}{t!} > \alpha\}.$

Let us recall the Gamma-Poisson relationship:
$P(X \leq x)_{X \sim {\rm Gamma}(\alpha,\beta)} = P(Y \geq \alpha)_{Y \sim {\rm Po}(x/\beta)},$
where $E[X]=\alpha\beta$ . (There are two different definitions for a Gamma distribution.)

We set $\alpha = T, \beta=2, x = 2n\lambda_0$ . (Then the distribution of $X$ is $\chi^2_{2T}$ ).

Now let $\psi_{2T}(\cdot)$ be the CDF of the chi-square distribution with $2T$ degrees of freedom.

Using $\psi_{2T}(\cdot)$ above confidence set can be rewritten as follows:
$C(\bm{x}) = \{\lambda_0 \mid \psi_{2T}(2n\lambda_0) > \alpha\} = \{\lambda_0 \mid \lambda_0 > \frac{1}{2n}\psi_{2T}^{-1}(\alpha)\} .$

And we conclude that $C(\bm{x}) = (\frac{1}{2n} \chi^2_{2T,1-\alpha}\infty)$ is the desired confidence set, where $\chi^2_{2T,1-\alpha}$ is the $\alpha$ th quantile of the chi-square distribution with $2T$ degree of freedom.