MATH 2411 Homework 3

Q1

i

df <- read.csv("lead_concentration.csv")
df_fall_1977 <- na.omit(df$Fall.1977)
 
n = length(df_fall_1977)
m = mean(df_fall_1977)
v = var(df_fall_1977)
s = sd(df_fall_1977)
 
alpha = (1 - 0.9) / 2
t_alpha = qt(1 - alpha, df=n-1)
 
c(m - t_alpha * s / sqrt(n), m + t_alpha * s / sqrt(n))

The 90% CI of mean is [8.504470, 9.858688]

ii

The CI for ${\sigma }^2$ is

$$\left[\frac{(n-1)S_{n-1}^2}{{\chi }_{n-1,\alpha }^2},\frac{(n-1)S_{n-1}^2}{{\chi }_{n-1,1-\alpha }^2}\right]$$

The CI for $\frac{1}{\sigma }$ is

$$\left[\sqrt{\frac{{\chi }_{n-1,1-\alpha }^2}{(n-1)S_{n-1}^2}},\sqrt{\frac{{\chi }_{n-1,\alpha }^2}{(n-1)S_{n-1}^2}}\right]$$

X_one_minus_alpha = qchisq(alpha, df=n-1)
X_alpha = qchisq(1 - alpha, df=n-1)
 
1 / sqrt(
    c(
        (n - 1) * v / X_one_minus_alpha,
        (n - 1) * v / X_alpha
    )
)

The 90% CI of $\frac{1}{\sigma }$ is [0.3260397, 0.4800552]

Q2

a

$$\begin{array}{rl}\mathrm{Bias}(\hat{p},p)& =E(\hat{p})-p\\ & =E(\bar{X})-p\\ & =E\left(\frac{Y}{n}\right)-p\\ & =E\left(\frac{X_1+X_2+\cdots +X_n}{n}\right)-p\\ & =\frac{1}{n}\sum _{i=1}^{n}E(X_i)-p\\ & =\frac{1}{n}np-p\\ & =0\end{array}$$

Therefore $\frac{Y}{n}$ is an unbiased estimator of $p$

b

$$\begin{array}{rl}Var(\bar{X})& =Var\left(\frac{X_1+X_2+\cdots +X_n}{n}\right)\\ & =\frac{1}{n^2}Var(X_1+X_2+\cdots +X_n)\\ & =\frac{1}{n^2}\left[Var(X_1)+Var(X_2)+\cdots +Var(X_n)\right]\\ & =\frac{np(1-p)}{n^2}\\ & =\frac{p(1-p)}{n}\end{array}$$

c

$$\begin{array}{rl}Var(\bar{X})=E({\bar{X}}^2)-E(\bar{X}{)}^2& =\frac{p(1-p)}{n}\\ E({\bar{X}}^2)& =\frac{p(1-p)}{n}+p^2\end{array}$$ $$\begin{array}{rl}E\left(\frac{\bar{X}(1-\bar{X})}{n}\right)& =E\left(\frac{\bar{X}-{\bar{X}}^2}{n}\right)\\ & =\frac{1}{n}(E(\bar{X})-E({\bar{X}}^2))\\ & =\frac{1}{n}\left(p-\left(\frac{p(1-p)}{n}+p^2\right)\right)\\ & =\frac{1}{n}\left(\frac{np-n{p}^2-p+p^2}{n}\right)\\ & =\frac{p^2(-n+1)+p(n-1)}{n^2}\\ & =\frac{(n-1)(-p^2+p)}{n^2}\\ & =\frac{(n-1)p(1-p)}{n^2}\end{array}$$

d

$$\begin{array}{rl}\text{Bias}=E(c\bar{X}(1-\bar{X}))-\frac{p(1-p)}{n}& =0\\ \frac{c(n-1)p(1-p)}{n}& =\frac{p(1-p)}{n}\\ c& =\frac{1}{n-1}\end{array}$$

$\text{pagebreak}$

Q3

i

The PDF of $X_i$ is

$$\begin{array}{r}p(x)=\frac{1}{T},0\le x\le T\end{array}$$

The expectation of $X_i$ is

$$\begin{array}{rl}E(X_i)& ={\int }_0^{T}xp(x)dx\\ & ={\int }_0^{T}x\frac{1}{T}dx\\ & =\frac{1}{T}\frac{T^2}{2}\\ & =\frac{T}{2}\end{array}$$

The expectation of $\bar{X}$ is

$$\begin{array}{rl}E(\bar{X})& =E\left(\frac{X_1+X_2+\cdots +X_n}{n}\right)\\ & =\frac{1}{n}E(X_1+X_2+\cdots +X_n)\\ & =\frac{1}{n}\times n\times \frac{T}{2}\\ & =\frac{T}{2}\end{array}$$

To make it unbiased, multiply by 2, then $\hat{T}=2\bar{X}$

$$\begin{array}{rl}Bias(\hat{T})=Bias(2\bar{X})& =E(2\bar{X})-T\\ & =T-T\\ & =0\end{array}$$

Therefore $\hat{T}=2\bar{X}$ is an unbiased estimator of $T$

ii

Since $T$ is the maximum time interval, $\max \left\{X_1,X_2,\dots ,X_n\right\}$ is the minimum value of $T$, which maximize $\frac{1}{T^n}$

$$\begin{array}{rl}{\hat{T}}_{MLE}& =\mathrm{argmax}\left\{p(X_1,X_2,\dots ,X_n)\right\}\\ & =\mathrm{argmax}\left\{\frac{1}{T^n}\right\}\\ & =\max \left\{X_1,X_2,\dots ,X_n\right\}\end{array}$$

iii

The CDF of ${\hat{T}}_{MLE}$ is

$$\begin{array}{rl}F(x)=P({\hat{T}}_{MLE}\le x)& =P(\max (X_1,X_2,\dots ,X_n)\le x)\\ & =P(X_1\le x)P(X_2\le x)\dots P(X_n\le x)\\ & ={\left(x\times \frac{1}{T}\right)}^n\\ & ={\left(\frac{x}{T}\right)}^n\end{array}$$

Differentiate w.r.t to $x$, the PDF of ${\hat{T}}_{MLE}$ is

$$\begin{array}{rl}f(x)& =\frac{1}{T^n}n{x}^{n-1}\\ & =\frac{n}{T^n}{x}^{n-1}\end{array}$$

$\text{pagebreak}$
The bias of ${\hat{T}}_{MLE}$ is

$$\begin{array}{rl}\mathrm{Bias}({\hat{T}}_{MLE})& =E({\hat{T}}_{MLE})-T\\ & ={\int }_0^{T}x\frac{n}{T^n}{x}^{n-1}dx-T\\ & =\frac{n}{T^n}{\left[\frac{x^{n+1}}{n+1}\right]}_0^T-T\\ & =\frac{n}{T^n}\frac{T^{n+1}}{n+1}-T\\ & =\frac{nT}{n+1}-T\ne 0\end{array}$$

Therefore ${\hat{T}}_{MLE}$ is a biased estimator

iv

Note that

$$\begin{array}{rl}{\sigma }_X^2=\mathrm{Var}(X)& =E(X^2)-[E(X){]}^2\\ & ={\int }_0^T{x}^2\frac{1}{T}dx-{\left(\frac{T}{2}\right)}^2\\ & =\frac{1}{T}{\left[\frac{x^3}{3}\right]}_0^T-\frac{T^2}{4}\\ & =\frac{T^2}{3}-\frac{T^2}{4}\\ & =\frac{T^2}{12}\end{array}$$

$\text{pagebreak}$

MSE of $\hat{T}$ is

$$\begin{array}{rl}MSE(\hat{T})& =(\mathrm{Bias}(\hat{T},T){)}^2+\mathrm{Var}(\hat{T})\\ & =0+\mathrm{Var}(2\bar{X})\\ & =4\mathrm{Var}(\bar{X})\\ & =4\times \frac{{\sigma }^2}{n}\\ & =4\times \frac{T^2}{12n}\\ & =\frac{T^2}{3n}\end{array}$$

Note that

$$\begin{array}{rl}\mathrm{Var}({\hat{T}}_{MLE})& =E(\max (X_1,X_2,\dots ,X_n{)}^2)-{\left[E(\max (X_1,X_2,\dots ,X_n))\right]}^2\\ & ={\int }_0^T{x}^2\frac{n}{T^n}{x}^{n-1}dx-{\left(\frac{nT}{n+1}\right)}^2\\ & =\frac{n}{T^n}{\left[\frac{x^{n+2}}{n+2}\right]}_0^T-{\left(\frac{nT}{n+1}\right)}^2\\ & =\frac{n{T}^2}{n+2}-{\left(\frac{nT}{n+1}\right)}^2\\ & =\frac{(n+1{)}^{2}n{T}^2-(n+2)n^2{T}^2}{(n+2)(n+1{)}^2}\\ & =\frac{n^3{T}^2+2n^2{T}^2+n{T}^2-n^3{T}^2-2n^2{T}^2}{(n+2)(n+1{)}^2}\\ & =\frac{n{T}^2}{(n+2)(n+1{)}^2}\end{array}$$

$\text{pagebreak}$

MSE of ${\hat{T}}_{MLE}$ is

$$\begin{array}{rl}MSE({\hat{T}}_{MLE})& =(\mathrm{Bias}({\hat{T}}_{MLE},T){)}^2+\mathrm{Var}({\hat{T}}_{MLE})\\ & ={\left(\frac{nT}{n+1}-T\right)}^2+\frac{n{T}^2}{(n+2)(n+1{)}^2}\\ & ={\left(\frac{nT-T(n+1)}{n+1}\right)}^2+\frac{n{T}^2}{(n+2)(n+1{)}^2}\\ & ={\left(\frac{nT-nT-T}{n+1}\right)}^2+\frac{n{T}^2}{(n+2)(n+1{)}^2}\\ & =\frac{T^2}{(n+1{)}^2}+\frac{n{T}^2}{(n+2)(n+1{)}^2}\\ & =\frac{n{T}^2+2T^2+n{T}^2}{(n+1{)}^2(n+2)}\\ & =\frac{2T^2(n+1)}{(n+1{)}^2(n+2)}\\ & =\frac{2T^2}{(n+1)(n+2)}\end{array}$$

Comparing the MSE of $\hat{T}$ and ${\hat{T}}_{MLE}$

$$\begin{array}{rl}\frac{MSE(\hat{T})}{MSE({\hat{T}}_{MLE})}& =\frac{n{T}^2}{(n+2)(n+1{)}^2}\frac{(n+1)(n+2)}{2T^2}\\ & =\frac{n}{2(n+1)}\\ & =\frac{n}{2n+2}\end{array}$$

$\frac{n}{2n+2}<1$ for all $n\ge 1$, therefore the MSE of ${\hat{T}}_{MLE}$ is larger