Computing probabilities with the ${\chi}^{2}_{n}$ distribution

Question

I have 2 questions. Here is background information to my questions, which involves showing that $\lbrace \sum_{i=1}^{n} x_{i} < \frac{\theta c_{1}}{2}\rbrace \cup \lbrace \sum_{i=1}^{n} x_{i} > \frac{\theta c_{2}}{2}\rbrace$ is the critical region for a UMPU level $\alpha$ test.

1. The random sample is coming from a $\operatorname{Exp}(\theta)$ distribution, $\theta>0$.

2. We can use the fact that if $X \sim \operatorname{Exp}(\theta)$ then $\frac{2X}{\theta} \sim \operatorname{Exp}(2) \sim {\chi}^{2}_{2}$.

3. We can further use the fact that if $Y_{i} \sim {\chi}^{2}_{2}$ then $\sum_{i=1}^{n} Y_{i} \sim {\chi}^{2}_{2n}$.

4. In this question, $f_{k}(x)$ is the pdf of a ${\chi}^{2}_{k}$ random variable.

Okay, I was able to find the region $\lbrace \sum_{i=1}^n x_i < \frac{\theta c_1} 2 \rbrace \cup \lbrace \sum_{i=1}^n x_i > \frac{\theta c_2} 2 \rbrace$ using the fact that $\operatorname{Exp}(\theta)$ has a montone likelihood ratio in $\sum x_i$ and using a theorem in class. That’s fine. All that is left to show is that $P(\lbrace \sum_{i=1}^n X_i < \frac{\theta c_1}{2}\rbrace \cup \lbrace \sum_{i=1}^n X_i > \frac{\theta c_2} 2\rbrace)=\alpha$.

My questions are:

1. Are the following steps valid, for $c_1<c_2$:

\begin{align} & P\left(\left\{ \sum_{i=1}^n X_i < \frac{\theta c_1} 2\right\} \cup \left\{ \sum_{i=1}^n X_i > \frac{\theta c_2} 2\right\}\right)=\alpha \\[10pt] \Longrightarrow & P(\chi^2_{2n}< c_1) + P(\chi^2_{2n}>c_2)=\alpha \\[10pt] \Longrightarrow & P(c_1<\chi^2_{2n}<c_2)=1-\alpha \\[10pt] \Longrightarrow & \int_{c_1}^{c_2} f_{2n}(x)dx=1-\alpha \\[10pt] \Longrightarrow & \int_{c_1}^{c_2} f_{2(n+1)}(x) \, dx=1-\alpha \end{align}

2. The equality $\int_{c_1}^{c_2} f_{2n}(x)\,dx=\int_{c_1}^{c_2} f_{2(n+1)}(x) \, dx$ was given. Why is this true?

Answer 1

Continued comment on distributions and confidence intervals: If $X_i$ are $n = 10$ random observations from an exponential population with mean $\mu,$ then $$\frac{\bar X}{\mu} \sim \mathsf{Gamma}(\text{shape} = 10, \text{rate} = 10).$$

Thus one can find $L$ and $U$ with $$0.95 = P(L < \bar X/\mu < U) = P(\bar X/U < \mu < \bar X/L),$$ so that $(\bar X/U,\, \bar X/L)$ is a 95% CI for $\mu.$ For convenience, one can use quantiles .025 and .975 for $L$ and $U$ of $\mathsf{Gamma}(\text{shape} = 10, \text{rate} = 10),$ respectively. [A shorter CI is possible by cutting different probabilities (that add to 5%) from the tails of the gamma distribution.]

Alternatively, using the chi-squared distribution with $df = 20,$ the same CI is of the form $(2T/U, 2T/L),$ where $T = \sum_i X_i = n\bar X$ and $L$ and $U$ are quantiles of $\mathsf{Chisq}(20).$

The computation for a sample of size ten from an exponential distribution with mean 3 (rate 1/3) in R statistical software is shown below. (In R, exponential and gamma distributions are parameterized according to the rate.)

set.seed(211)  # retain this line to use same sample; delete for different sample
n = 10;  mu = 3;  lam = 1/3     # lam is 'rate'
x = rexp(n, lam);  a = mean(x)  # sample of size n; and its mean
a; a/qgamma(c(.975,.025), n, n)
## 3.435613                     # numerical value of mean of sample
## 2.010918 7.164410            # 95% CI based on sample
2*10*a/qchisq(c(.975,.025), 2*n)
## 2.010918 7.164410            # same CI using CHISQ(2n)

Answer 2

The answer to my first question is that it’s correct.

The answer to my second question is no, that equality is not true. The equality is derived from a condition, based on the theorem that I used, to compute $c_{1}$ and $c_{2}$, and it’s specific to that certain problem. In general, the equality is not true.