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

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}

1. 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?
• If I understand your notation correctly, your inequality is not correct. Use 10 and 12 degrees of freedom and find the probability of the interval $(.5, 5)$ then R code a = .5; b = 5; diff(pchisq(c(a,b), 10)) returns 0.1088154 and diff(pchisq(c(a,b), 12)) returns 0.04202076. // Maybe this is intended to be an approximation for a specific case. If you will give specific numbers, I'll look at it again. // Many books use chi-sq to find exponential and gamma probabilities because traditionally chi-sq was the only related dist'n tabled. With modern software, best to seek direct answers. Feb 11, 2018 at 16:26
• Thanks Bruce. I guess my real question was, if the following was true, for $c_{1}<c_{2}$: $P({\chi^{2}}_{2n}< c_{1}) + P({\chi^{2}}_{2n}>c_{2})=\alpha$ $\Rightarrow P(c_{1}<{\chi^{2}}_{2n}<c_{2})=1-\alpha$ Feb 11, 2018 at 17:08
• That's correct. Specifically, suppose $X\sim\mathsf{Chisq}(df=10).$ Then $P(X < 3.27) = P(X > 20.48)$ $= 0.025.$ In R, qchisq(c(.025,.975), 10) returns 3.246973 and 20.483177. So, $P(3.27 < X < 20.48) = .95:$ diff(pchisq(c(3.27, 20.48),10)) returns 0.9493112. In R, pchisq is a chi-sq CDF and qchisq is a chi-sq quantile function (inverse CDF). // You should be able to approximate the results from R by looking at a table of chi-sq with 10 df. Feb 11, 2018 at 19:01
• To make a 95% confidence interval, most practitioners are happy enough to cut 2.5% from each tail, obtaining what is called a 'probability symmetric' CI. Not necessarily the shortest possible CI. However, for your UMPU test you want the sum of the tail probabilities to be 5% (for a test at 5% level), but generally want different probabilities in the tails (adding to 5%) chosen to give the greatest power. // Do you know the specific values of $c_1$ and $c_2$ in your problem (based on a numerical value of $\bar X$) or is this a purely theoretical problem? Feb 11, 2018 at 19:08
• There were no specific values given. It’s a graduate statistical inference course, so it’s all just theoretical work. Thanks again though for your help. Feb 11, 2018 at 19:19

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)


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.