8.3.4 Solved Problems
Let $X_1$, $X_2$, $X_3$, $...$, $X_n$ be a random sample from an exponential distribution with parameter $\theta$, i.e., \begin{align} f_{X_i}(x;\theta) = \theta e^{-\theta x}u(x). \end{align} Our goal is to find a $(1-\alpha)100\%$ confidence interval for $\theta$. To do this, we need to remember a few facts about the gamma distribution. More specifically, If $Y=X_1+X_2+\cdots+X_n$, where the $X_i$'s are independent $Exponential(\theta)$ random variables, then $Y \sim Gamma(n, \theta)$. Thus, the random variable $Q$ defined as \begin{equation} Q=\theta (X_1+X_2+\cdots+X_n) \end{equation} has a $Gamma(n, 1)$ distribution. Let us define $\gamma_{p,n}$ as follows. For any $p \in [0,1]$ and $n \in \mathbb{N}$, we define $\gamma_{p,n}$ as the real value for which \begin{align}%\label{} P(Q > \gamma_{p,n})=p, \end{align} where $Q \sim Gamma(n, 1)$.
- Explain why $Q=\theta (X_1+X_2+\cdots+X_n)$ is a pivotal quantity.
- Using $Q$ and the definition of $\gamma_{p,n}$, construct a $(1-\alpha)100\%$ confidence interval for $\theta$.
- Solution
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- $Q$ is a function of the $X_i$'s and $\theta$, and its distribution does not depend on $\theta$ or any other unknown parameters. Thus, $Q$ is a pivotal quantity.
- Using the definition of $\gamma_{p,n}$, a $(1-\alpha)$ interval for $Q$ can be stated as \begin{align}%\label{} P\left(\gamma_{1-\frac{\alpha}{2},n-1} \leq Q \leq \gamma_{\frac{\alpha}{2},n-1} \right)= 1-\alpha. \end{align} Therefore, \begin{align}%\label{} P\left(\gamma_{1-\frac{\alpha}{2},n-1} \leq \theta (X_1+X_2+\cdots+X_n) \leq \gamma_{\frac{\alpha}{2},n-1} \right)= 1-\alpha. \end{align} Since $X_1+X_2+\cdots+X_n$ is always a positive quantity, the above equation is equivalent to \begin{align}%\label{} P\left(\frac{\gamma_{1-\frac{\alpha}{2},n-1}}{X_1+X_2+\cdots+X_n} \leq \theta \leq \frac{\gamma_{\frac{\alpha}{2},n-1}}{X_1+X_2+\cdots+X_n} \right)= 1-\alpha. \end{align} We conclude that $\left[\frac{\gamma_{1-\frac{\alpha}{2},n-1}}{X_1+X_2+\cdots+X_n}, \frac{\gamma_{\frac{\alpha}{2},n-1}}{X_1+X_2+\cdots+X_n}\right]$ is a $(1-\alpha)100\%$ confidence interval for $\theta$.
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Problem
A random sample $X_1$, $X_2$, $X_3$, $...$, $X_{100}$ is given from a distribution with known variance $\textrm{Var}(X_i)=16$. For the observed sample, the sample mean is $\overline{X}=23.5$. Find an approximate $95 \%$ confidence interval for $\theta=EX_i$.
- Solution
- Here, $\left[\overline{X}- z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}} , \overline{X}+ z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}}\right]$ is an approximate $(1-\alpha)100\%$ confidence interval. Since $\alpha=0.05$, we have \begin{align}%\label{} z_{\frac{\alpha}{2}}=z_{0.025}=\Phi^{-1}(1-0.025)=1.96 \end{align} Also, $\sigma=4$. Therefore, the approximate confidence interval is \begin{align} \left[23.5- 1.96 \frac{4}{\sqrt{100}} , 23.5- 1.96 \frac{4}{\sqrt{100}}\right] \approx [ 22.7 , 24.3 ]. \end{align}
Problem
To estimate the portion of voters who plan to vote for Candidate A in an election, a random sample of size $n$ from the voters is chosen. The sampling is done with replacement. Let $\theta$ be the portion of voters who plan to vote for Candidate A among all voters. How large does $n$ need to be so that we can obtain a $90 \%$ confidence interval with $3 \%$ margin of error? That is, how large $n$ needs to be such that \begin{align}%\label{} P\left(\overline{X}-0.03 \leq \theta \leq \overline{X}+ 0.03 \right) \geq 0.90, \end{align} where $\overline{X}$ is the portion of people in our random sample that say they plan to vote for Candidate A.
- Solution
- Here, \begin{align}%\label{} \left[\overline{X}- \frac{z_{\frac{\alpha}{2}}}{2\sqrt{n}} , \overline{X}+ \frac{z_{\frac{\alpha}{2}}}{2\sqrt{n}}\right] \end{align} is an approximate $(1-\alpha)100\%$ confidence interval for $\theta$. Since $\alpha=0.1$, we have \begin{align}%\label{} z_{\frac{\alpha}{2}}=z_{0.05}=\Phi^{-1}(1-0.05)=1.645 \end{align} Therefore, we need to have \begin{align}%\label{} \frac{1.645}{2\sqrt{n}}=0.03 \end{align} Therefore, we obtain \begin{align}%\label{} n=\left(\frac{1.645}{2 \times 0.03}\right)^2. \end{align} We conclude $n \geq 752$ is enough.
Problem
- Let $X$ be a random variable such that $R_X \subset [a,b]$, i.e., we always have $a \leq X \leq b$. Show that \begin{equation} \textrm{Var}(X) \leq \frac{(b-a)^2}{4}. \end{equation}
- Let $X_1$, $X_2$, $X_3$, $...$, $X_n$ be a random sample from an unknown distribution with CDF $F_X(x)$ such that $R_X \subset [a,b]$. Specifically, $EX$ and $\textrm{Var}(X)$ are unknown. Find a $(1-\alpha)100\%$ confidence interval for $\theta=EX$. Assume that $n$ is large.
- Solution
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- Define $Y=X-\frac{a+b}{2}$. Thus, $R_Y \subset [-\frac{b-a}{2}, \frac{b-a}{2}]$. Then, \begin{align} \textrm{Var}(X) &=\textrm{Var} (Y)\\ &=E[Y^2]-\mu_Y^2\\ &\leq E[Y^2]\\ &\leq \left(\frac{b-a}{2}\right)^2 \quad \left(\textrm{since } Y^2 \leq \left(\frac{b-a}{2}\right)^2\right)\\ &=\frac{(b-a)^2}{4}. \end{align}
- Here, we have an upper bound on $\sigma$, which is $\sigma_{max}=\frac{(b-a)}{2}$. Thus, the interval \begin{align}%\label{} \left[\overline{X}- z_{\frac{\alpha}{2}} \frac{\sigma_{max}}{\sqrt{n}} , \overline{X}+ z_{\frac{\alpha}{2}} \frac{\sigma_{max}}{\sqrt{n}}\right] \end{align} is a $(1-\alpha)100\%$ confidence interval for $\theta$. More specifically, \begin{align}%\label{} \left[\overline{X}- z_{\frac{\alpha}{2}} \frac{b-a}{2\sqrt{n}} , \overline{X}+ z_{\frac{\alpha}{2}} \frac{b-a}{2\sqrt{n}} \right] \end{align} is a $(1-\alpha)100\%$ confidence interval for $\theta$.
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Problem
A random sample $X_1$, $X_2$, $X_3$, $...$, $X_{144}$ is given from a distribution with unknown variance $\textrm{Var}(X_i)=\sigma^2$. For the observed sample, the sample mean is $\overline{X}=55.2$, and the sample variance is $S^2=34.5$. Find a $99 \%$ confidence interval for $\theta=EX_i$.
- Solution
- The interval \begin{equation} \left[\overline{X}- z_{\frac{\alpha}{2}}\frac{S}{\sqrt{n}} , \overline{X}+ z_{\frac{\alpha}{2}} \frac{S}{\sqrt{n}}\right] \end{equation} is approximately a $(1-\alpha)100\%$ confidence interval for $\theta$. Here, $n=144$, $\alpha=0.01$, so we need \begin{align}%\label{} z_{\frac{\alpha}{2}}=z_{0.005}=\Phi^{-1}(1-0.005) \approx 2.58 \end{align} Thus, we can obtain a $99 \%$ confidence interval for $\theta$ as \begin{align}%\label{} \left[\overline{X}- z_{\frac{\alpha}{2}}\frac{S}{\sqrt{n}} , \overline{X}+ z_{\frac{\alpha}{2}} \frac{S}{\sqrt{n}}\right] &=\left[55.2-2.58 \cdot \frac{\sqrt{34.5}}{12}, 55.2+2.58 \cdot \frac{\sqrt{34.5}}{12}\right]\\ &\approx[53.94, 56.46]. \end{align} Therefore, $[53.94, 56.46]$ is an approximate $99 \%$ confidence interval for $\theta$.
Problem
A random sample $X_1$, $X_2$, $X_3$, $...$, $X_{16}$ is given from a normal distribution with unknown mean $\mu=EX_i$ and unknown variance $\textrm{Var}(X_i)=\sigma^2$. For the observed sample, the sample mean is $\overline{X}=16.7$, and the sample variance is $S^2=7.5$.
- Find a $95 \%$ confidence interval for $\mu$.
- Find a $95 \%$ confidence interval for $\sigma^2$.
- Solution
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- Here, the interval \begin{align} \left[\overline{X}-t_{\frac{\alpha}{2},n-1}\frac{S}{\sqrt{n}} , \overline{X}+ t_{\frac{\alpha}{2},n-1} \frac{S}{\sqrt{n}}\right] \end{align} is a $(1-\alpha)100\%$ confidence interval for $\mu$. Let $n=16$, $\alpha=0.05$, then \begin{align}%\label{} t_{0.025,15} \approx 2.13 \end{align} The above value can obtained in MATLAB using the command $\mathtt{tinv(0.975,15)}$. Thus, we can obtain a $95 \%$ confidence interval for $\mu$ as \begin{align}%\label{} \left[16.7- 2.13 \frac{\sqrt{7.5}}{4} , 16.7+ 2.13 \frac{\sqrt{7.5}}{4}\right] \approx [15.24 , 18.16]. \end{align} Therefore, $[15.24 , 18.16]$ is a $95 \%$ confidence interval for $\mu$.
- Here, $\left[\frac{(n-1)S^2}{\chi^2_{\frac{\alpha}{2},n-1}} , \frac{(n-1)S^2}{\chi^2_{1-\frac{\alpha}{2},n-1}} \right]$ is a $(1-\alpha)100\%$ confidence interval for $\sigma^2$. In this problem, $n=16$, $\alpha=.05$, so we need \begin{align}%\label{} \chi^2_{0.025,15} \approx 27.49, \quad \chi^2_{0.975,15} \approx 6.26 \end{align} The above values can obtained in MATLAB using the commands $\mathtt{chi2inv(0.975,15)}$ and $\mathtt{chi2inv(0.025,15)}$, respectively. Thus, we can obtain a $95 \%$ confidence interval for $\sigma^2$ as \begin{align}%\label{} \left[\frac{(n-1)S^2}{\chi^2_{\frac{\alpha}{2},n-1}} , \frac{(n-1)S^2}{\chi^2_{1-\frac{\alpha}{2},n-1}} \right] &=\left[\frac{15 \times 7.5}{27.49} , \frac{15 \times 7.5}{6.26} \right]\\ &\approx [4.09 , 17.97]. \end{align} Therefore, $[4.09 , 17.97]$ is a $95 \%$ confidence interval for $\sigma^2$.
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