Exercises from 5th ed Book

Chapter 4

4.59

Let $Y_1<Y_2<Y_3<Y_4<Y_5$ denote the order statistics of a random sample of size 5 from a distribution having p.d.f. $f(x)=e^{-x}$, $0<x<\infty$, zero elsewhere. Show that $Z_1=Y_2$ and $Z_2=Y_4-Y_2$ are independent.

Hint: First find the joint p.d.f. of $Y_2$ and $Y_4$.

Answer

Diketahui $f(x)=e^{-x}$ untuk $x>0$, merupakan distribusi exponential dengan parameter $\lambda =1$.

cdf dari distribusi eksponensial adalah: $F(x)=1-e^{-x},x>0$

Berdasarkan teorema marginal pdf of order statistics, joint marginal pdf dari $Y_i$ dan $Y_j$ dengan $i<j$ adalah:

$$g_{ij}\left(y_i,y_j\right)=\{\begin{array}{ll}\frac{n!}{(i-1)!(j-i-1)!(n-j)!}{\left[F\left(y_i\right)\right]}^{i-1}{\left[F\left(y_j\right)-F\left(y_i\right)\right]}^{j-i-1}{\left[1-F\left(y_j\right)\right]}^{n-j}f\left(y_i\right)f\left(y_j\right)& a<y_i<y_j<b\\ 0& \text{ elsewhere }\end{array}$$

Untuk $Y_2$ dan $Y_4$ dengan $n=5$, $i=2$, $j=4$:

$$\begin{array}{rl}{g}_{2,4}(y_2,y_4)& =\frac{5!}{(2-1)!(4-2-1)!(5-4)!}[F(y_2){]}^{2-1}[F(y_4)-F(y_2){]}^{4-2-1}[1-F(y_4){]}^{5-4}f(y_2)f(y_4)\\ & =120(1-e^{-u})(e^{-u}-e^{-v})(e^{-v})(e^{-u})(e^{-v})\\ & =120(1-e^{-u})(e^{-u}-e^{-v})e^{-u-2v}\end{array}$$

Misalkan $z_1=u$ dan $z_2=v-u$. Dapat diperoleh

$$\begin{array}{rl}{g}_{2,4}(y_2,y_4)& =120(1-e^{-z_1})(e^{-z_1}-e^{-(z_1+z_2)})e^{-z_1-2(z_1+z_2)}\\ & =120(1-e^{-z_1})(e^{-z_1}-e^{-z_1}{e}^{-z_2})e^{-3z_1-2z_2}\\ & =120(1-e^{-z_1})e^{-z_1}(1-e^{-z_2})e^{-3z_1-2z_2}\\ & =120(1-e^{-z_1})e^{-4z_1}(1-e^{-z_2})e^{-2z_2}\\ & =[120(1-e^{-z_1})e^{-4z_1}]\cdot [(1-e^{-z_2})e^{-2z_2}]\end{array}$$

Karena joint pdf $g_{2,4}(y_2,y_4)$ dapat diekspresikan sebagai $g(z_1)\cdot h(z_2)$, maka $Z_1$ dan $Z_2$ adalah independen.

$\therefore$ Terbukti bahwa $Z_1=Y_2$ dan $Z_2=Y_4-Y_2$ adalah independen.

Chapter 5

5.7

Let $Y_n$ : Sequence of random variable, with $Y_n\sim b(n,p)$ (check here)

Prove that $1-Y_n/n$ converges in probability to $1-p$

Answer

Karena $Y_n\sim b(n,p)$, berarti $E[Y_n]=np$ dan $\mathrm{Var}(Y_n)=np(1-p)$. Misalkan ${\bar{Y}}_n=\frac{Y_n}{n}$. Maka,

• $E[\overline{Y_n}]=\frac{E[Y_n]}{n}=\frac{np}{n}=p$
• $\text{Var}(\overline{Y_n})=\frac{\text{Var}(Y_n)}{n^2}=\frac{np(1-p)}{n^2}=\frac{p(1-p)}{n}$

Misalkan $k=\frac{ϵ}{\sigma }$. Berdasarkan teorema Chebyshev,

$$\begin{array}{rl}P\left(\left|{\bar{Y}}_n-E[X]\right|>k\sigma \right)& \le \frac{1}{k^2}\\ P\left(\left|\frac{Y_n}{n}-p\right|>ϵ\right)& \le \frac{{\sigma }^2}{{ϵ}^2}\\ & \le \frac{\mathrm{Var}({\bar{Y}}_n)}{{ϵ}^2}=\frac{p(1-p)}{n{ϵ}^2}\\ \underset{n\to \infty }{\lim }P\left(\left|\frac{Y_n}{n}-p\right|>ϵ\right)& \le \underset{n\to \infty }{\lim }\frac{p(1-p)}{n{ϵ}^2}=0\end{array}$$

Sehingga diperoleh

$$\underset{n\to \infty }{\lim }P\left(\left|1-\frac{Y_n}{n}-(1-p)\right|>ϵ\right)=0$$

Berdasarkan definisi konvergen dalam probabilitas, terbukti bahwa $1-\frac{Y_n}{n}\stackrel{P}{\to }1-p$

5.8

Let $S_n^2$ denote the variance of a random sample of size $n$ from a distribution that is $N(\mu ,{\sigma }^2)$. Prove that $n{S}_n^2/(n-1)$ converges in probability to ${\sigma }^2$.

Answer

Diketahui $S_n^2=\frac{1}{n-1}{\sum }_{i=1}^n(X_i-\bar{X}{)}^2$ adalah varians sampel dari distribusi $N(\mu ,{\sigma }^2)$ dengan $S_n^2\sim N\left({\sigma }^2,\frac{2{\sigma }^4}{n-1}\right)$.

Misalkan $T_n=\frac{n{S}_n^2}{n-1}$. Maka,

$$\begin{array}{rl}E[T_n]& =E\left[\frac{n{S}_n^2}{n-1}\right]\\ & =\frac{n}{n-1}E[S_n^2]\\ & =\frac{n}{n-1}{\sigma }^2\\ \text{Var}(T_n)& =\text{Var}\left(\frac{n{S}_n^2}{n-1}\right)\\ & ={\left(\frac{n}{n-1}\right)}^2\text{Var}(S_n^2)\\ & ={\left(\frac{n}{n-1}\right)}^2\frac{2{\sigma }^4}{n-1}\\ & =\frac{2n^2{\sigma }^4}{(n-1{)}^3}.\end{array}$$

Berdasarkan teorema Chebyshev, untuk setiap $ϵ>0$ berlaku

$$P\left(\left|T_n-{\sigma }^2\right|>ϵ\right)\le \frac{\text{Var}(T_n)}{{ϵ}^2}=\frac{2n^2{\sigma }^4}{(n-1{)}^3{ϵ}^2}.$$

Karena ${\lim }_{n\to \infty }\frac{2n^2{\sigma }^4}{(n-1{)}^3{ϵ}^2}=0$, maka:

$$\underset{n\to \infty }{\lim }P\left(\left|T_n-{\sigma }^2\right|>ϵ\right)=0.$$

Berdasarkan definisi konvergen dalam probabilitas, terbukti bahwa $\frac{n{S}_n^2}{n-1}\stackrel{P}{\to }{\sigma }^2$.

$\therefore$ Terbukti bahwa $n{S}_n^2/(n-1)$ konvergen dalam probabilitas ke ${\sigma }^2$.

5.10

Let $Y_n$ : $n$-th Order statistic from random sample of size $n$, with $Y_n\sim U(0,\theta )$

Prove that $Z_n=\sqrt{Y_n}$ converges in probability to $\sqrt{\theta }$

Answer

Dari Example 1 Section 5.1

$Y_n$ converges in distribution to a random variable that has a degenerate distribution at the point $y=\theta$.

Artinya, $Y_n\stackrel{D}{\to }\theta$. Berdasarkan teorema 5.2.2, diperoleh $Y_n\stackrel{P}{\to }\theta$. Misalkan $Z_n=g(Y_n)=\sqrt{Y_n}$. Berdasarkan teorema 5.1.4, diperoleh $Z_n\stackrel{P}{\to }\theta$, sehingga terbukti bahwa $Z_n=\sqrt{Y_n}$ konvergen dalam probabilitas ke $\sqrt{\theta }$

5.11

Let $X_n$ have a gamma distribution with parameter $\alpha =n$ and $\beta$, where $\beta$ is not a function of $n$. Let $Y_n=X_n/n$. Find the limiting distribution of $Y_n$.

Answer

Diketahui $X_n\sim \Gamma (n,\beta )$, memiliki mgf $M_{X_n}(t)=(1-\beta t{)}^{-n}$, untuk $t<\frac{1}{\beta }$.

Misalkan $Y_n=\frac{X_n}{n}$. mgf dari $Y_n$ adalah:

$$\begin{array}{rl}{M}_{Y_n}(t)& =E[e^{t{Y}_n}]\\ & =E\left[e^{t\cdot \frac{X_n}{n}}\right]\\ & =M_{X_n}\left(\frac{t}{n}\right)\\ & ={\left(1-\beta \frac{t}{n}\right)}^{-n}\\ & ={\left(1-\frac{\beta t}{n}\right)}^{-n}.\end{array}$$

Karena ${\lim }_{n\to \infty }{\left(1+\frac{b}{n}\right)}^{cn}=e^{bc}$, maka ${\lim }_{n\to \infty }{M}_{Y_n}(t)=e^{\beta t}$

Misalkan $X$ adalah variabel acak dengan mgf $M_X(t)=e^{\beta t}$. $X$ adalah mgf dari distribusi degenerate di $\beta$, yaitu $P(X=\beta )=1$. Berdasarkan teorema teknik mgf, karena ${\lim }_{n\to \infty }{M}_{Y_n}(t)=M_X(t)$, maka $Y_n\stackrel{D}{\to }X$.

$\therefore$ Distribusi limit dari $Y_n$ adalah distribusi degenerate di $\beta$.

5.12

Let

• $Z_n$ : Sequence of random variables, with $Z_n\sim {\chi }^2(n)$
• $W_n=Z_n/n^2$

Find the limiting distribution of $W_n$

Answer

Diketahui mgf dari distribusi Chi-square adalah $M_{Z_n}(t)=(1-2t{)}^{-n/2}$, untuk $t<\frac{1}{2}$. Karena $W_n=\frac{Z_n}{n^2}$, mgf dari $W_n$ dapat diperoleh dengan

$$\begin{array}{rl}{M}_{W_t}& =E[e^{t{W}_n}]\\ & =E\left[e^{\frac{t}{n^2}{Z}_n}\right]\\ & =M_{Z_n}\left(\frac{t}{n^2}\right)\\ & ={\left(1-\frac{2t}{n^2}\right)}^{-n/2},\text{Untuk }\frac{t}{n^2}<\frac{1}{2}\end{array}$$

Karena ${\lim }_{n\to \infty }{\left(1+\frac{b}{n}\right)}^{cn}=e^{bc}$, dengan memisalkan $b=-2t$, $c=1$, dan $n=n^2$, dapat diperoleh

$$\begin{array}{rl}{M}_{W_n}(t)& ={\left(1-\frac{2t}{n^2}\right)}^{-n/2}\\ & ={\left[{\left(1-\frac{2t}{n^2}\right)}^{n^2}\right]}^{-1/(2n)}\\ \underset{n\to \infty }{\lim }{M}_{W_n}(t)& =\underset{n\to \infty }{\lim }{\left(e^{-2t}\right)}^{-1/(2n)}\\ & =\underset{n\to \infty }{\lim }{e}^{t/n}\\ & =e^0\\ & =1\end{array}$$

Misalkan $X$ variabel acak dengan mgf $M_X(t)$. Andaikan $X$ berdistribusi degenerate ke $0$, yang berakibat $M_X(t)=E\left[e^{tX}\right]=e^{t\cdot 0}=1$. Karena ${\lim }_{n\to \infty }{M}_{W_n}(t)=M_X(t)$, berdasarkan teorema 5.2.10, diperoleh $W_n\stackrel{D}{\to }X$. Terbukti bahwa limiting distribution dari $W_n$ adalah $X$.

5.13

Let $X$ : Random variable, with $X\sim {\chi }^2(50)$

Approximate $\mathrm{Pr}(40<X<60)$

Answer

5.15

Let

• $Z_n$ : Sequence of random variable, with $Z_n\sim P(n)$
• $Y_n=(Z_n-n)/\sqrt{n}$

Show that the limiting distribution of $Y_n$ is normal with mean zero and variance $1$.

Answer

Diketahui mgf dari distribusi $P(n)$ adalah $M_{Z_n}=e^{n(e^t-1)}$. Dapat diperoleh mgf dari $M_{Y_n}$:

$$\begin{array}{rl}{M}_{Y_n}(t)& =E\left[e^{t{Y}_n}\right]\\ & =E\left[e^{t\cdot \frac{Z_n-n}{\sqrt{n}}}\right]\\ & =e^{-t\sqrt{n}}E\left[e^{t/\sqrt{n}{Z}_n}\right]\\ & =e^{-t\sqrt{n}}{M}_{Z_n}\left(\frac{t}{\sqrt{n}}\right)\\ & =e^{-t\sqrt{n}}{e}^{n(e^{t/\sqrt{n}}-1)}\end{array}$$

Menggunakan Maclaurun series pada $f(x)=e^x$, perluaskan $e^{t/\sqrt{n}}$ menjadi

$$e^{t/\sqrt{n}}=1+\frac{t}{\sqrt{n}}+\frac{t^2}{2n}+o\left(\frac{1}{n}\right)=\sum _{n=0}^{\infty }\frac{(t/\sqrt{n}{)}^n}{n!}$$

Sehingga diperoleh

$$\begin{array}{rl}{M}_{Y_n}(t)& =e^{-t\sqrt{n}}{e}^{n(e^{t/\sqrt{n}}-1)}\\ & =e^{-t\sqrt{n}}{e}^{n\left(t/\sqrt{n}+t^2/2n+o(1/n)\right)}\\ & =e^{-t\sqrt{n}}{e}^{t\sqrt{n}+t^2/2+o(1)}\\ & =e^{t^2/2+o(1)}\\ \underset{n\to \infty }{\lim }{M}_{Y_n}(t)& =\underset{n\to \infty }{\lim }{e}^{t^2/2+o(1)}\\ & =e^{t^2/2}\end{array}$$

Misalkan $X$ variabel acak dengan berdistribusi $N(0,1)$. Artinya, $M_X(t)=e^{t^2/2}$. Berdasarkan teorema 5.2.10, karena ${\lim }_{n\to \infty }{M}_{Y_n}(t)=M_X(t)$, maka $Y_n\stackrel{D}{\to }X$, sehingga $Y_n\stackrel{D}{\to }N(0,1)$. Terbukti bahwa limiting distribution dari $Y_n$ adalah distribusi normal dengan mean $0$ dan variansi $1$.

Chapter 6

6.23

Let $\bar{X}$ denote the mean of a random sample of size 25 from a gamma-type distribution with $\alpha =4$ and $\beta >0$. Use the central limit theorem to find an approximate 0.954 confidence Interval for $\mu$, the mean of the gamma distribution.

Hint: Base the confidence interval on the random variable $(\bar{X}-4\beta )/(4{\beta }^2/25{)}^{1/2}=5\bar{X}/2\beta -10$

Answer Untuk distribusi gamma dengan parameter $\alpha =4$ dan $\beta$, diketahui:

$$\begin{array}{rl}E[X]& =\alpha \beta =4\beta =\mu \\ \text{Var}(X)& =\alpha {\beta }^2=4{\beta }^2\end{array}$$

Berdasarkan teorema limit pusat, untuk sampel berukuran $n=25$:

$$\begin{array}{rl}\frac{\bar{X}-\mu }{\sqrt{\text{Var}(\bar{X})}}& =\frac{\bar{X}-4\beta }{\sqrt{4{\beta }^2/25}}\\ & =\frac{\bar{X}-4\beta }{2\beta /5}\\ & =\frac{5\bar{X}}{2\beta }-10\stackrel{D}{\to }N(0,1)\end{array}$$

Karena diketahui $\mu =4\beta$, untuk confidence interval 95.4%, dengan $z_{0.023}=2$, dapat diperoleh:

$$\begin{array}{rl}0.954& =P\left(-2<\frac{5\bar{X}}{2\beta }-10<2\right)\\ & =P\left(-2<\frac{5\bar{X}}{2\beta }-10<2\right)\\ & =P\left(8<\frac{5\bar{X}}{2\beta }<12\right)\\ & =P\left(\frac{16\beta }{5}<\bar{X}<\frac{24\beta }{5}\right)\\ & =P\left(\frac{5\bar{X}}{24}<\beta <\frac{5\bar{X}}{16}\right)\\ & =P\left(\frac{5\bar{X}}{6}<4\beta <\frac{5\bar{X}}{4}\right)\\ & =P\left(\frac{5\bar{X}}{6}<\mu <\frac{5\bar{X}}{4}\right)\end{array}$$

$\therefore$ Confidence interval 95.4% untuk $\mu$ adalah $\boxed{\left(\frac{5\bar{X}}{6},\frac{5\bar{X}}{4}\right)}$

6.26

It is known that a random variable $X$ has a Poisson distribution with parameter $\mu$. A sample of 200 observations from this population has a mean equal to 3.4. Construct an approximate 90 percent confidence Interval for $\mu$.

Answer

Untuk distribusi Poisson dengan parameter $\mu$, diketahui:

$$\begin{array}{rl}E[X]& =\mu \\ \text{Var}(X)& =\mu \end{array}$$

Berdasarkan teorema limit pusat, untuk sampel berukuran besar $n=200$: $\frac{\bar{X}-\mu }{\sqrt{\mu /n}}\stackrel{D}{\to }N(0,1)$

Untuk confidence interval 90%, dengan $\alpha =0.10$ sehingga $z_{0.05}=1.645$, dapat diperoleh: $P\left(-1.645<\frac{\bar{X}-\mu }{\sqrt{\mu /n}}<1.645\right)=0.90$

Dengan $\bar{x}=3.4$ dan $n=200$, kita perlu menyelesaikan:

$$\begin{array}{rl}-1.645& <\frac{3.4-\mu }{\sqrt{\mu /200}}<1.645\\ -1.645\sqrt{\frac{\mu }{200}}& <3.4-\mu <1.645\sqrt{\frac{\mu }{200}}\end{array}$$

Karena $n$ besar, dapat diaproksimasikan dengan $\sqrt{\mu /n}\approx \sqrt{\bar{x}/n}=\sqrt{3.4/200}=\sqrt{0.017}=0.1304$

Sehingga:

$$\begin{array}{rl}-1.645(0.1304)& <3.4-\mu <1.645(0.1304)\\ -0.2145& <3.4-\mu <0.2145\\ 3.4-0.2145& <\mu <3.4+0.2145\\ 3.1855& <\mu <3.6145\end{array}$$

$\therefore$ Confidence interval 90% untuk $\mu$ adalah $\boxed{(3.185,3.615)}$

6.27

Let $Y_1<Y_2<\cdots <Y_n$ denote the order statistics of a random sample of size $n$ from a distribution that has p.d.f. $f(x)=3x^2/{\theta }^3,0<x<\theta$, zero elsewhere.

(a) Show that $\mathrm{Pr}(c<Y_n/\theta <1)=1-c^{3n}$, where $0<c<1$ (b) If $n$ is 4 and if the observed value of $Y$, is 2.3, what is a 95 percent confidence interval for 8?

Answer

6.27.a

Diketahui cdf dari $X$:

$F(x)={\int }_0^x\frac{3t^2}{{\theta }^3}dt=\frac{x^3}{{\theta }^3},0<x<\theta$

Berdasarkan CDF of order statistics, maksimum $Y_n$ distribusinya adalah: $F_{Y_n}(y)=[F(y){]}^n={\left(\frac{y^3}{{\theta }^3}\right)}^n=\frac{y^{3n}}{{\theta }^{3n}}$

Maka:

$$\begin{array}{rl}P(c<Y_n/\theta <1)& =P(c\theta <Y_n<\theta )\\ & =F_{Y_n}(\theta )-F_{Y_n}(c\theta )\\ & =\frac{{\theta }^{3n}}{{\theta }^{3n}}-\frac{(c\theta {)}^{3n}}{{\theta }^{3n}}\\ & =\boxed{1-c^{3n}}\end{array}$$

$\therefore$ Terbukti bahwa $\mathrm{Pr}(c<Y_n/\theta <1)=1-c^{3n}$

6.27.b

Untuk $n=4$ dan confidence itnerval 95%, berarti:

$$\begin{array}{rl}P(c<Y_4/\theta <1)& =1-c^{12}=0.95\\ ⟺c^{12}& =0.05\\ ⟺c& =(0.05{)}^{1/12}\\ & =0.7943\end{array}$$

Karena $P(c<Y_4/\theta <1)=0.95$ dan $Y_4=2.3$, dapat peroleh:

$$\begin{array}{rl}& \\ c& <\frac{Y_4}{\theta }<1\\ \frac{Y_4}{1}& <\theta <\frac{Y_4}{c}\\ Y_4& <\theta <\frac{Y_4}{0.7943}\\ 2.3& <\theta <\frac{2.3}{0.7943}\\ 2.3& <\theta <2.896\end{array}$$

$\therefore$ Confidence interval 95% untuk $\theta$ adalah $\boxed{(2.3,2.896)}$

6.29

Let $X_1,X_2,\dots ,X_n$ be a random sample from a gamma distribution with known parameter $\alpha =3$ and unknown $\beta >0$. Discuss the construction of a confidence interval for $\beta$.

Hint: What is the distribution of $2{\sum }_{i=1}^n{X}_i/\beta$? Follow the procedure outlined in Exercise 6.28.

Answer

Dari Gamma Distribution Relationships, diketahui bahwa jika $X_i\sim \text{Gamma}(\alpha ,\beta )$, maka: $\frac{2X_i}{\beta }\sim {\chi }^2(2\alpha )$

Sehingga untuk $\alpha =3$: $\frac{2X_i}{\beta }\sim {\chi }^2(6)$

Karena $X_1,X_2,\dots ,X_n$ saling bebas, maka:

$$\begin{array}{rl}\frac{2{\sum }_{i=1}^n{X}_i}{\beta }& =\sum _{i=1}^n\frac{2X_i}{\beta }\\ & \sim {\chi }^2(6n)\end{array}$$

Untuk konstruksi confidence interval $(1-\alpha )100\%$ :

$$\begin{array}{rl}1-\alpha & =P\left({\chi }_{\alpha /2,6n}^2<\frac{2{\sum }_{i=1}^n{X}_i}{\beta }<{\chi }_{1-\alpha /2,6n}^2\right)\\ & =P\left({\chi }_{\alpha /2,6n}^2<\frac{2{\sum }_{i=1}^n{X}_i}{\beta }<{\chi }_{1-\alpha /2,6n}^2\right)\\ & =\boxed{P\left(\frac{2{\sum }_{i=1}^n{X}_i}{{\chi }_{1-\alpha /2,6n}^2}<\beta <\frac{2{\sum }_{i=1}^n{X}_i}{{\chi }_{\alpha /2,6n}^2}\right)}\end{array}$$

$\therefore$ Confidence interval $(1-\alpha )100\%$ untuk $\beta$ adalah $\boxed{\left(\frac{2{\sum }_{i=1}^n{X}_i}{{\chi }_{1-\alpha /2,6n}^2},\frac{2{\sum }_{i=1}^n{X}_i}{{\chi }_{\alpha /2,6n}^2}\right)}$

6.35

Let $X$ and $Y$ be the means of two independent random sample, each of size $n$, from the respective distributions $N({\mu }_1,{\sigma }^2)$ and $N({\mu }_2,{\sigma }^2)$, where the common variance is known. Find $n$ such that

$\mathrm{Pr}(\bar{X}-\bar{Y}-\sigma /5<{\mu }_1-{\mu }_2<\bar{X}-\bar{Y}+\sigma /5)=0.90$

Answer

Karena $\bar{X}\sim N({\mu }_1,{\sigma }^2/n)$ dan $\bar{Y}\sim N({\mu }_2,{\sigma }^2/n)$ saling bebas, maka:

$$\begin{array}{rl}E[\bar{X}-\bar{Y}]& ={\mu }_1-{\mu }_2\\ & \\ \text{Var}(\bar{X}-\bar{Y})& =\text{Var}(\bar{X})+\text{Var}(\bar{Y})\\ & =\frac{{\sigma }^2}{n}+\frac{{\sigma }^2}{n}\\ & =\frac{2{\sigma }^2}{n}\end{array}$$

Sehingga $\bar{X}-\bar{Y}\sim N\left({\mu }_1-{\mu }_2,\frac{2{\sigma }^2}{n}\right)$

Perhatikan bahwa $\frac{(\bar{X}-\bar{Y})-({\mu }_1-{\mu }_2)}{\sqrt{2{\sigma }^2/n}}=\frac{(\bar{X}-\bar{Y})-({\mu }_1-{\mu }_2)}{\sigma \sqrt{2/n}}\sim N(0,1)$

Sehingga

$$\begin{array}{rl}0.90& =P(\bar{X}-\bar{Y}-\sigma /5<{\mu }_1-{\mu }_2<\bar{X}-\bar{Y}+\sigma /5)\\ & =P(-\sigma /5<(\bar{X}-\bar{Y})-({\mu }_1-{\mu }_2)<\sigma /5)\\ & =P\left(\frac{-\sigma /5}{\sigma \sqrt{2/n}}<\frac{(\bar{X}-\bar{Y})-({\mu }_1-{\mu }_2)}{\sigma \sqrt{2/n}}<\frac{\sigma /5}{\sigma \sqrt{2/n}}\right)\\ & =P\left(\frac{-1/5}{\sqrt{2/n}}<Z<\frac{1/5}{\sqrt{2/n}}\right)\\ & =P\left(-\frac{\sqrt{n}}{5\sqrt{2}}<Z<\frac{\sqrt{n}}{5\sqrt{2}}\right)\end{array}$$

Untuk confidence interval 90%, $z_{0.05}=1.645$. Sehingga:

$$\begin{array}{rl}\frac{\sqrt{n}}{5\sqrt{2}}& =1.645\\ \sqrt{n}& =1.645\times 5\sqrt{2}\\ \sqrt{n}& =8.225\sqrt{2}\\ \sqrt{n}& =11.632\\ n& =135.3\end{array}$$

$\therefore$ Nilai $n$ yang diperlukan adalah $\boxed{n=136}$ (dibulatkan ke atas)

Chapter 7

7.40

Let

• $X_1,X_2,\dots ,X_n$ : Random sample, with
  • distribution $N(\theta ,1),-\infty <\theta <\infty$

Find MVUE of ${\theta }^2$

Hint: First determine $E({\bar{X}}^2)$

Answer

Diketahui $\mathrm{Var}(X_i)=1$, dan $\bar{X}=\frac{1}{n}{\sum }_{i=1}^n{X}_i$ adalah statistik tak bias untuk $\theta$. Maka $E(\bar{X})=\theta$. Perhatikan bahwa

$$\begin{array}{rl}\mathrm{Var}(\bar{X})& =\mathrm{Var}\left[\left(\frac{1}{n}\sum _{i=1}^n{X}_i\right)\right]\\ & =\frac{1}{n^2}\sum _{i=1}^n\mathrm{Var}(X_i)\\ & =\frac{1}{n^2}\sum _{i=1}^{n}1\\ & =\frac{1}{n^2}\cdot n\\ & =\frac{1}{n}\end{array}$$

Sehingga dapat diperoleh

$$\begin{array}{rl}\mathrm{Var}(\bar{X})& =E({\bar{X}}^2)-[E(\bar{X}){]}^2\\ ⟺E({\bar{X}}^2)& =\mathrm{Var}(\bar{X})+[E(\bar{X}){]}^2\\ & =\frac{1}{n}+{\theta }^2\end{array}$$

Misalkan $Y={\bar{X}}^2-1/n$. Maka, dapat diperoleh

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

Diketahui $\bar{X}$ adalah statistik cukup yang komplit. Ingat bahwa $Y$ adalah fungsi dari $\bar{X}$, sehingga berdasarkan teorema Lehmann-Scheffé, $Y={\bar{X}}^2-1/n$ adalah MVUE untuk ${\theta }^2$.

$\therefore$ MVUE dari ${\theta }^2$ adalah $\boxed{{\bar{X}}^2-1/n}$

7.41

Let $X_1,\dots ,X_n$ : Random sample, with distribution $N(0,\theta )$.

Then $Y=\sum X_i^2$ is a complete sufficient statistic for $\theta$.

Find MVUE of ${\theta }^2$

Answer

Karena $X_i\sim N(0,\theta )$, dapat diperoleh

$$\begin{array}{rl}& X_i\sim N(0,\theta )\\ ⟺& \frac{X_i}{\sqrt{\theta }}\sim N(0,1)\\ ⟺& \frac{X_i^2}{\theta }\sim {\chi }^2(1)\\ ⟺& \sum _{i=1}^n\frac{X_i^2}{\theta }\sim {\chi }^2(n)\\ ⟺& \frac{Y}{\theta }\sim {\chi }^2(n)\end{array}$$

Jadi,

• $E\left[\frac{Y}{\theta }\right]=n$
• $\text{Var}\left(\frac{Y}{\theta }\right)=2n⟺\mathrm{Var}(Y)={\theta }^2\cdot 2n$

Perhatikan bahwa

$$\begin{array}{rl}E[Y^2]& =\text{Var}(Y)+(E[Y]{)}^2\\ & =2n{\theta }^2+n^2{\theta }^2\\ & =(n^2+2n){\theta }^2\\ ⟺E\left[\frac{Y^2}{n^2+2n}\right]& ={\theta }^2\end{array}$$

Karena $\frac{Y^2}{n^2+2n}$ adalah fungsi dari statistik cukup yang komplit $Y$, berdasarkan teorema Lehmann-Scheffé, $\frac{Y^2}{n^2+2n}$ adalah MVUE untuk ${\theta }^2$.

$\therefore$ MVUE dari ${\theta }^2$ adalah $\boxed{\frac{Y^2}{n^2+2n}}$

7.42

In the notation of Example 2 of this section, is there an UMVE of $\mathrm{Pr}(-c\le X\le c)$? Here $c>0$.

Answer

Untuk $X\sim N(\theta ,1)$ dapat diperoleh $\mathrm{Pr}(-c\le X\le c)=\mathrm{Pr}(X\le c)-\mathrm{Pr}(X\le -c)=\Phi (c-\theta )-\Phi (-c-\theta )$

Misalkan

$$u(X_1)=\{\begin{array}{ll}1& \text{if }-c\le X_1\le c\\ 0& \text{lainnya}\end{array}$$

Maka, $E[u(X_1)]=\mathrm{Pr}(-c\le X_1\le c)=\Phi (c-\theta )-\Phi (-c-\theta )$

Berdasarkan teorema Rao-Blackwell dan Lehmann and Scheffe Theorem, $\phi (\bar{X})=E[u(X_1)|\bar{X}=\bar{x}]$ adalah UMVUE.

Diberikan $\bar{X}=\bar{x}$, $X_1$ memiliki distribusi kondisional, $N\left(\theta +\frac{\rho {\sigma }_1}{{\sigma }_2}(\bar{x}-\theta ),{\sigma }_1^2(1-{\rho }^2)\right)$ dimana:

• $\rho =\frac{1}{\sqrt{n}}$ (koefisien korelasi)
• ${\sigma }_1^2=1$, ${\sigma }_2^2=\frac{1}{n}$

Sehingga diperoleh $X_1|\bar{X}=\bar{x}\sim N\left(\bar{x},\frac{n-1}{n}\right)$. Akibatnya,

$$\begin{array}{rl}\phi (\bar{x})& =\mathrm{Pr}(-c\le X_1\le c|\bar{X}=\bar{x})\\ & =\mathrm{Pr}\left(\frac{-c-\bar{x}}{\sqrt{\frac{n-1}{n}}}\le \frac{X_1-\bar{x}}{\sqrt{\frac{n-1}{n}}}\le \frac{c-\bar{x}}{\sqrt{\frac{n-1}{n}}}\right)\\ & =\Phi \left(\sqrt{\frac{n}{n-1}}\cdot (c-\bar{x})\right)-\Phi \left(\sqrt{\frac{n}{n-1}}\cdot (-c-\bar{x})\right)\\ & =\Phi \left(\sqrt{\frac{n}{n-1}}\cdot (c-\bar{x})\right)-\Phi \left(-\sqrt{\frac{n}{n-1}}\cdot (c+\bar{x})\right)\end{array}$$

$\therefore$ Jadi, UMVUE dari $\mathrm{Pr}(-c\le X\le c)$ adalah $\boxed{\Phi \left(\sqrt{\frac{n}{n-1}}\cdot (c-\bar{X})\right)-\Phi \left(-\sqrt{\frac{n}{n-1}}\cdot (c+\bar{X})\right)}$

Chapter 8

8.15

Let $X$ have a Gamma distribution with $\alpha =4$ and $\beta =\theta >0$. a. Find the Fisher Information $I(\theta )$. b. If $X_1,X_2,\dots ,X_n$ is a random sample from this distribution, show that the m.l.e. of $\theta$ is an efficient estimator of $\theta$.

8.15.a

Diketahui pdf dari distribusi Gamma dengan $\alpha =4$ dan $\beta =\theta$ adalah: $f(x;\theta )=\frac{1}{\Gamma (4){\theta }^4}{x}^{4-1}{e}^{-x/\theta }=\frac{1}{6{\theta }^4}{x}^3{e}^{-x/\theta }$

Fungsi log-likelihood untuk satu pengamatan $X$ adalah:

$$\begin{array}{rl}\ln L(\theta )& =\ln \left(\frac{1}{6{\theta }^4}{x}^3{e}^{-x/\theta }\right)\\ & =\ln (1)-\ln (6{\theta }^4)+\ln (x^3)-\frac{x}{\theta }\\ & =-\ln (6)-4\ln (\theta )+3\ln (x)-\frac{x}{\theta }\end{array}$$

Turunan pertama $\ln L()\theta$: $\frac{\partial \ln L}{\partial \theta }=\frac{\partial }{\partial \theta }\left(-\ln (6)-4\ln \theta +3\ln x-\frac{x}{\theta }\right)=-\frac{4}{\theta }+\frac{x}{{\theta }^2}$

Turunan kedua $\ln L(\theta )$: $\frac{{\partial }^2\ln L}{\partial {\theta }^2}=\frac{\partial }{\partial \theta }\left(-\frac{4}{\theta }+\frac{x}{{\theta }^2}\right)=\frac{4}{{\theta }^2}-\frac{2x}{{\theta }^3}$

Sehingga diperoleh

$$\begin{array}{rl}I(\theta )& =-E\left[\frac{{\partial }^2\ln L}{\partial {\theta }^2}\right]\\ & =-E\left[\frac{4}{{\theta }^2}-\frac{2X}{{\theta }^3}\right]\\ & =-\left(\frac{4}{{\theta }^2}-\frac{2E[X]}{{\theta }^3}\right)\\ & =-\frac{4}{{\theta }^2}+\frac{2(4\theta )}{{\theta }^3}\\ & =-\frac{4}{{\theta }^2}+\frac{8\theta }{{\theta }^3}\\ & =\boxed{\frac{4}{{\theta }^2}}\end{array}$$

8.15.b

Untuk sampel acak ukuran $n$, log-likelihoodnya adalah: $\ln L(\theta )={\sum }_{i=1}^n\ln f(x_i;\theta )={\sum }_{i=1}^n\left(-\ln 6-4\ln \theta +3\ln x_i-\frac{x_i}{\theta }\right)=-n\ln 6-4n\ln \theta +3{\sum }_{i=1}^n\ln x_i-\frac{1}{\theta }{\sum }_{i=1}^n{x}_i$

Sehingga didapati

$$\begin{array}{rl}\frac{\partial \ln L}{\partial \theta }& =-\frac{4n}{\theta }+\frac{1}{{\theta }^2}\sum _{i=1}^n{x}_i=0\\ & ⟺\frac{1}{{\theta }^2}\sum _{i=1}^n{x}_i=\frac{4n}{\theta }\\ & ⟺\hat{\theta }=\frac{\sum X_i}{4n}=\frac{\bar{X}}{4}\end{array}$$

Perhatikan bahwa

$$\begin{array}{rl}E[\hat{\theta }]& =E\left[\frac{\bar{X}}{4}\right]\\ & =\frac{1}{4}E[\bar{X}]\\ & =\frac{1}{4}E[X]\\ & =\frac{1}{4}(4\theta )\\ & =\theta .\end{array}$$

Artinya, $\hat{\theta }$ adalaha penaksir tak bias untuk $\theta$

Ingat bahwa $X\sim \Gamma (4,\theta )$. Diketahui $\mathrm{Var}(X)=4{\theta }^2$. Dapat diperoleh

$$\begin{array}{rl}Var(\hat{\theta })& =Var\left(\frac{\bar{X}}{4}\right)\\ & =\frac{1}{16}Var(\bar{X})\\ & =\frac{1}{16}\frac{Var(X)}{n}\\ & =\frac{1}{16}\frac{4{\theta }^2}{n}\\ & =\frac{{\theta }^2}{4n}\end{array}$$

Rao-Cramer Lower Bound untuk penaksir tak bias dari $\theta$ adalah: $\frac{1}{nI(\theta )}=\frac{1}{n(4/{\theta }^2)}=\frac{{\theta }^2}{4n}$

Karena $\mathrm{Var}(\hat{\theta })=\frac{1}{nI(\theta )}$, maka berdasarkan Corollary 6.2.1 Rao-Cramér bound for unbiased estimators, $\frac{{\theta }^2}{4n}$ adalah penaksir efisien untuk $\theta$.

8.27

If $X_1,X_2,\dots ,X_n$ is a random sample from a distribution with p.d.f. $f(x;\theta )=3{\theta }^3(x+\theta {)}^{-4}$, $0<x<\theta$, zero elsewhere, where $0<\theta$, show that $Y=2X$ is an unbiased estimator of $\theta$ and determine its efficiency.

Jawab

Perhatikan bahwa

$$\begin{array}{rl}E(X)& ={\int }_0^{\infty }x\cdot 3{\theta }^3(x+\theta {)}^{-4}dx\\ & =3{\theta }^3{\int }_{\theta }^{\infty }(u-\theta )u^{-4}du\\ & =3{\theta }^3{\int }_{\theta }^{\infty }(u^{-3}-\theta u^{-4})du\\ & =3{\theta }^3{\left[-\frac{u^{-2}}{2}+\frac{\theta u^{-3}}{3}\right]}_{\theta }^{\infty }\\ & =3{\theta }^3\left(\frac{1}{2{\theta }^2}-\frac{1}{3{\theta }^2}\right)\\ & =\frac{\theta }{2}\end{array}$$

sehingga

$$\begin{array}{rl}E(Y)& =E(2X)\\ & =2E(X)\\ & =2\cdot \frac{\theta }{2}\\ & =\theta \end{array}$$

sehingga terbukti bahwa Y adalah penaksir tak bias untuk $\theta$

Selanjutnya, akan dicari efisiensi dari $Y$. Pertama, akan dicari variansi $Y$:

$$\begin{array}{rl}E[X^2]& ={\int }_0^{\infty }{x}^2\cdot 3{\theta }^3(x+\theta {)}^{-4}dx\\ & =3{\theta }^3{\int }_{\theta }^{\infty }(u-\theta {)}^2{u}^{-4}du\\ & =3{\theta }^3{\int }_{\theta }^{\infty }(u^{-2}-2\theta u^{-3}+{\theta }^2{u}^{-4})du\\ & ={\theta }^2\\ & \\ \mathrm{Var}(X)& =E[X^2]-(E[X]{)}^2\\ & ={\theta }^2-(\theta /2{)}^2\\ & =\frac{3{\theta }^2}{4}\\ & \\ \mathrm{Var}(Y)& =\mathrm{Var}(2\bar{X})\\ & =4\mathrm{Var}(\bar{X})\\ & =\frac{4}{n}\mathrm{Var}(X)\\ & =\frac{3{\theta }^2}{n}\end{array}$$

Informasi Fisher $I(\theta )$: $\ln f(x;\theta )=\ln 3+3\ln \theta -4\ln (x+\theta )⟹\frac{{\partial }^2\ln f}{\partial {\theta }^2}=-\frac{3}{{\theta }^2}+\frac{4}{(x+\theta {)}^2}$

$I(\theta )=-E\left[-\frac{3}{{\theta }^2}+\frac{4}{(X+\theta {)}^2}\right]=\frac{3}{{\theta }^2}-4E\left[\frac{1}{(X+\theta {)}^2}\right]$ $E\left[\frac{1}{(X+\theta {)}^2}\right]={\int }_0^{\infty }\frac{3{\theta }^3}{(x+\theta {)}^6}dx=\frac{3}{5{\theta }^2}$ $I(\theta )=\frac{3}{{\theta }^2}-\frac{12}{5{\theta }^2}=\frac{3}{5{\theta }^2}$

$$\begin{array}{rl}\frac{[k^{\prime }(\theta ){]}^2/nI(\theta )}{\mathrm{Var}(Y)}& =\frac{1/n\cdot \frac{3}{5{\theta }^2}}{3{\theta }^2/n}(k^{\prime }(\theta )=1\text{ karena }Y\text{ tak bias})\\ & =\boxed{\frac{5}{9}}\end{array}$$

8.29

If $X_1,X_2,\dots ,X_n$ is a random sample from $N(\theta ,1)$, find a lower bound of the variance of an estimator of $k(\theta )={\theta }^2$. Determine an unbiased minimum variance estimator of ${\theta }^2$ and then compute its efficiency.

Jawab

Untuk $N(\theta ,1)$, diketahui $I(\theta )=1$. Batas bawah Cramer-Rao untuk $k(\theta )={\theta }^2$ adalah: $\frac{[k^{\prime }(\theta ){]}^2}{nI(\theta )}=\frac{(2\theta {)}^2}{n(1)}=\frac{4{\theta }^2}{n}$

Diketahui $\bar{X}\sim N(\theta ,1/n)$ adalah statistik cukup yang komplit untuk $\theta$

Perhatikan bahwa

$$\begin{array}{rl}E[{\bar{X}}^2]& =\mathrm{Var}(\bar{X})+(E[\bar{X}]{)}^2\\ & =\frac{1}{n}+{\theta }^2\end{array}$$

Sehingga, $E[{\bar{X}}^2-1/n]={\theta }^2$. Artinya, $T={\bar{X}}^2-1/n$ adalah penaksir tak bias untuk ${\theta }^2$. Berdasarkan teorema Lehmann Scheffe, $T$ adalah MVUE untuk ${\theta }^2$.

Selanjutnya, akan dicari variansi dari $T$: $\mathrm{Var}(T)=\mathrm{Var}({\bar{X}}^2-1/n)=\mathrm{Var}({\bar{X}}^2)$

Misal $Z=\sqrt{n}(\bar{X}-\theta )\sim N(0,1)$, maka $\bar{X}=Z/\sqrt{n}+\theta$. Sehingga ${\bar{X}}^2={\left(\frac{Z}{\sqrt{n}}+\theta \right)}^2=\frac{Z^2}{n}+\frac{2\theta Z}{\sqrt{n}}+{\theta }^2$ Dan variansinya adalah

$$\begin{array}{rl}\mathrm{Var}({\bar{X}}^2)& =\mathrm{Var}\left(\frac{Z^2}{n}+\frac{2\theta Z}{\sqrt{n}}+{\theta }^2\right)\\ & =\mathrm{Var}\left(\frac{Z^2}{n}+\frac{2\theta Z}{\sqrt{n}}\right)\\ & =\frac{1}{n^2}\mathrm{Var}(Z^2)+\frac{4{\theta }^2}{n}\mathrm{Var}(Z)+\frac{4\theta }{n\sqrt{n}}\mathrm{Cov}(Z^2,Z)\end{array}$$

Untuk $Z\sim N(0,1)$, diketahui $\mathrm{Var}(Z)=1$, $\mathrm{Var}(Z^2)=2$, dan $\mathrm{Cov}(Z^2,Z)=0$. Maka: $\mathrm{Var}(T)=\frac{2}{n^2}+\frac{4{\theta }^2}{n}-2\cdot 0$

Efisiensi dari $T$ adalah perbandingan antara batas bawah cramer (CRLB) dan variansi $T$:

$$\begin{array}{rl}\text{Efisiensi}& =\frac{\text{CRLB}}{\mathrm{Var}(T)}\\ & =\frac{4{\theta }^2/n}{2/n^2+4{\theta }^2/n}\\ & =\frac{4{\theta }^2/n}{(2+4n{\theta }^2)/n^2}\\ & =\frac{4n{\theta }^2}{2+4n{\theta }^2}\\ & =\boxed{\frac{2n{\theta }^2}{1+2n{\theta }^2}}\end{array}$$

Chapter 9

9.6

Let $X_1,X_2,\dots ,X_{10}$ be a random sample from a distribution that is $N({\theta }_1,{\theta }_2)$¹. Find a best test of the simple hypothesis $H_0:{\theta }_1={\theta }_1^{\prime }=0,{\theta }_2={\theta }_2^{\prime }=1$ against the alternative simple hypothesis $H_1:{\theta }_1={\theta }_1^{\prime \prime }=1,{\theta }_2={\theta }_2^{\prime \prime }=4$.

Answer

Untuk $H_0:{\theta }_1=0,{\theta }_2=1$, fungsi likelihood adalah:

$$\begin{array}{rl}{L}_0(x_1,\dots ,x_{10})& =\prod _{i=1}^{10}\frac{1}{\sqrt{2\pi \cdot 1}}\exp \left(-\frac{(x_i-0{)}^2}{2\cdot 1}\right)\\ & =(2\pi {)}^{-5}\exp \left(-\frac{1}{2}\sum _{i=1}^{10}{x}_i^2\right)\end{array}$$

Untuk $H_1:{\theta }_1=1,{\theta }_2=4$, fungsi likelihood adalah:

$$\begin{array}{rl}{L}_1(x_1,\dots ,x_{10})& =\prod _{i=1}^{10}\frac{1}{\sqrt{2\pi \cdot 4}}\exp \left(-\frac{(x_i-1{)}^2}{2\cdot 4}\right)\\ & =(8\pi {)}^{-5}\exp \left(-\frac{1}{8}\sum _{i=1}^{10}(x_i-1{)}^2\right)\end{array}$$

Berdasarkan Neyman-Pearson Theorem, best critical region didefinisikan oleh:

$$\begin{array}{rl}\frac{L_1}{L_0}& \ge k\\ \frac{(8\pi {)}^{-5}\exp \left(-\frac{1}{8}{\sum }_{i=1}^{10}(x_i-1{)}^2\right)}{(2\pi {)}^{-5}\exp \left(-\frac{1}{2}{\sum }_{i=1}^{10}{x}_i^2\right)}& \ge k\\ \frac{(2\pi {)}^{-5}}{(8\pi {)}^{-5}}\exp \left(-\frac{1}{8}\sum _{i=1}^{10}(x_i-1{)}^2+\frac{1}{2}\sum _{i=1}^{10}{x}_i^2\right)& \ge k\\ \frac{1}{4^5}\exp \left(-\frac{1}{8}\sum _{i=1}^{10}(x_i-1{)}^2+\frac{1}{2}\sum _{i=1}^{10}{x}_i^2\right)& \ge k\end{array}$$

Perhatikan bahwa:

$$\begin{array}{rl}-\frac{1}{8}\sum _{i=1}^{10}(x_i-1{)}^2+\frac{1}{2}\sum _{i=1}^{10}{x}_i^2& =-\frac{1}{8}\sum _{i=1}^{10}(x_i^2-2x_i+1)+\frac{1}{2}\sum _{i=1}^{10}{x}_i^2\\ & =-\frac{1}{8}\sum _{i=1}^{10}{x}_i^2+\frac{1}{4}\sum _{i=1}^{10}{x}_i-\frac{10}{8}+\frac{1}{2}\sum _{i=1}^{10}{x}_i^2\\ & =\frac{3}{8}\sum _{i=1}^{10}{x}_i^2+\frac{1}{4}\sum _{i=1}^{10}{x}_i-\frac{5}{4}\end{array}$$

Ingat bahwa $k$ adalah suatu konstanta positif. Misalkan $k^{\prime }=\ln (k)+5\ln (4)+\frac{5}{4}$. Dapat diperoleh:

$$\begin{array}{rl}\frac{L_1}{L_0}=\frac{1}{4^5}\exp \left(\frac{3}{8}\sum _{i=1}^{10}{x}_i^2+\frac{1}{4}\sum _{i=1}^{10}{x}_i-\frac{5}{4}\right)& \ge k\\ \frac{3}{8}\sum _{i=1}^{10}{x}_i^2+\frac{1}{4}\sum _{i=1}^{10}{x}_i-\frac{5}{4}& \ge \ln (k\cdot 4^5)\\ 3\sum _{i=1}^{10}{x}_i^2+2\sum _{i=1}^{10}{x}_i& \ge k^{\prime }\end{array}$$

Karena koefisien dari $\sum x_i^2$ dan $\sum x_i$ keduanya positif, critical region dapat ditulis sebagai:

$\boxed{C:3\sum _{i=1}^{10}{X}_i^2+2\sum _{i=1}^{10}{X}_i\ge k^{\prime }}$

untuk suatu konstanta $k$ positif dengan $k^{\prime }=\ln (k)+5\ln (4)+\frac{5}{4}$.

$\therefore$ Best test untuk hipotesis yang diberikan adalah tolak $H_0$ jika $3{\sum }_{i=1}^{10}{X}_i^2+2{\sum }_{i=1}^{10}{X}_i\ge k^{\prime }$.

9.10

Let $X_1,X_2,\dots ,X_{10}$ denote a random sample of size $10$ from a Poisson distribution with mean $\theta$. Show that the critical region $C$ defined by ${\sum }_{i=1}^{10}{x}_i\ge 3$ is a best critical region for testing $H_0:\theta =0.1$ against $H_1:\theta =0.5$. Determine, for this test, the significance level $\alpha$ and power at $\theta =0.5$.

Jawab

Diketahui pmf dari distribusi Poisson adalah $f(x;\theta )=\frac{{\theta }^x{e}^{-\theta }}{x!},x=0,1,2,\dots$

Fungsi likelihood untuk sampel $X_1,X_2,\dots ,X_{10}$ adalah:

$$\begin{array}{rl}L(\theta )& =\prod _{i=1}^{10}\frac{{\theta }^{x_i}{e}^{-\theta }}{x_i!}\\ & =\frac{{\theta }^{{\sum }_{i=1}^{10}{x}_i}{e}^{-10\theta }}{{\prod }_{i=1}^{10}{x}_i!}\end{array}$$

Untuk $H_0:\theta =0.1$: $L_0=\frac{(0.1{)}^{\sum x_i}{e}^{-1}}{\prod x_i!}$

Untuk $H_1:\theta =0.5$: $L_1=\frac{(0.5{)}^{\sum x_i}{e}^{-5}}{\prod x_i!}$

Sehingga diperoleh

$$\begin{array}{rl}\frac{L_1}{L_0}& =\frac{(0.5{)}^{\sum x_i}{e}^{-5}}{(0.1{)}^{\sum x_i}{e}^{-1}}\\ & ={\left(\frac{0.5}{0.1}\right)}^{\sum x_i}{e}^{-4}\\ & =5^{\sum x_i}{e}^{-4}\end{array}$$

Berdasarkan Neyman-Pearson Theorem, best critical region didefinisikan oleh:

$$\begin{array}{rl}\frac{L_1}{L_0}& \ge k\\ 5^{\sum x_i}{e}^{-4}& \ge k\\ 5^{\sum x_i}& \ge k{e}^4\\ \sum x_i& \ge {\log }_5(k{e}^4)\end{array}$$

Karena ${\log }_5(k{e}^4)$ adalah konstanta positif critical region berbentuk ${\sum }_{i=1}^{10}{x}_i\ge c$ untuk suatu konstanta $c$, artinya, $C:{\sum }_{i=1}^{10}{x}_i\ge 3$ adalah best critical region.

Di bawah $H_0$, ${\sum }_{i=1}^{10}{X}_i\sim \text{Poisson}(10\times 0.1)=\text{Poisson}(1)$ sehingga

$$\begin{array}{rl}\alpha & =P_{H_0}\left[\sum _{i=1}^{10}{X}_i\ge 3\right]\\ & =1-P[\sum _{i=1}^{10}{X}_i\le 2]\\ & =1-\left(P[Y=0]+P[Y=1]+P[Y=2]\right)\\ & =1-\left(\frac{1^0{e}^{-1}}{0!}+\frac{1^1{e}^{-1}}{1!}+\frac{1^2{e}^{-1}}{2!}\right)\\ & =1-e^{-1}\left(1+1+\frac{1}{2}\right)\\ & =1-\frac{5e^{-1}}{2}\\ & =\boxed{1-\frac{5}{2e}\approx 0.0803}\end{array}$$

Di bawah $H_1$, ${\sum }_{i=1}^{10}{X}_i\sim \text{Poisson}(10\times 0.5)=\text{Poisson}(5)$. Misal $Y\sim \text{Poisson(5)}$,sehingga

$$\begin{array}{rl}\alpha & =P_{H_1}\left[\sum _{i=1}^{10}{X}_i\ge 3\right]\\ & =1-P[Y\le 2]\\ & =1-\left(\frac{5^0{e}^{-5}}{0!}+\frac{5^1{e}^{-5}}{1!}+\frac{5^2{e}^{-5}}{2!}\right)\\ & =1-e^{-5}\left(1+5+\frac{25}{2}\right)\\ & =1-\frac{37e^{-5}}{2}\\ & =\boxed{1-\frac{37}{2e^5}\approx 0.875}\end{array}$$

$\therefore$ Critical region $C:{\sum }_{i=1}^{10}{x}_i\ge 3$ adalah best critical region dengan significane level $\alpha \approx 0.0803$ dan power $\approx 0.875$

9.21

Let $X_1,X_2,\dots ,X_n$ be a random sample from a distribution with pdf $f(x;\theta )=\theta x^{\theta -1},0<x<\infty$, zero elsewhere, where $\theta >0$. Find a sufficient statistic for $\theta$ and show that a uniformly most powerful test of $H_0:\theta =6$ against $H_1:\theta <6$ is based on this statistic.

Jawab

Fungsi likelihood untuk sampel adalah:

$$\begin{array}{rl}L(\theta )& =\prod _{i=1}^n\theta x_i^{\theta -1}\\ & ={\theta }^n\prod _{i=1}^n{x}_i^{\theta -1}\\ & ={\theta }^n{\left(\prod _{i=1}^n{x}_i\right)}^{\theta -1}\end{array}$$

Log-likelihood:

$$\begin{array}{rl}\ln L(\theta )& =n\ln \theta +(\theta -1)\sum _{i=1}^n\ln x_i\\ & =n\ln \theta +(\theta -1)\ln \left(\prod _{i=1}^n{x}_i\right)\end{array}$$

Note

Untuk mencari UMPT, perlu dicari UMP Critical Region, perlu dicari Best Critical Region menggunakan Neyman-Pearson Theorem. Selain itu, dari Hogg&Craig ed8,

Perhatikan bahwa berdasarkan Neyman Theorem, $Y_1={\prod }_{i=1}^n{X}_i=u_1(\mathbf{X})$ adalah statistik cukup untuk $\theta$ dengan:

$$\begin{array}{rl}Y& ={\theta }^n{\left(\prod x_i\right)}^{\theta -1}\cdot 1\\ & =k_1[u_1(\mathbf{x});\theta ]\cdot k_2(\mathbf{x})\end{array}$$

Misalkan ${\theta }_1<6$. Rasio likelihood adalah:

$$\begin{array}{rl}\frac{L({\theta }_1)}{L(6)}& =\frac{{\theta }_1^n{\left(\prod x_i\right)}^{{\theta }_1-1}}{6^n{\left(\prod x_i\right)}^{6-1}}\\ & ={\left(\frac{{\theta }_1}{6}\right)}^n{\left(\prod x_i\right)}^{{\theta }_1-6}\end{array}$$

Rasio log:

$$\begin{array}{rl}\ln \left(\frac{L({\theta }_1)}{L(6)}\right)& =n\ln \left(\frac{{\theta }_1}{6}\right)+({\theta }_1-6)\ln \left(\prod x_i\right)\\ & =n\ln \left(\frac{{\theta }_1}{6}\right)+({\theta }_1-6)\sum _{i=1}^n\ln x_i\end{array}$$

Lalu:

1. Karena ${\theta }_1<6$, maka ${\theta }_1-6<0$.
2. Karena $\ln x_i$ negatif untuk $0<x_i<1$, maka $({\theta }_1-6)\sum \ln x_i>0$
3. Perhatikan bahwa $({\theta }_1-6)\sum \ln x_i$ meningkat ketika $\sum \ln x_i$ semakin negatif
4. Karena $\sum \ln x_i=\ln \left(\prod x_i\right)$ maka $\ln \left(\frac{L({\theta }_1)}{L(6)}\right)$ meningkat ketika $\prod x_i$ mengecil ($y$ mengecil maka $\ln (y)$ membesar)
5. Sehingga, $\forall {\theta }_1<6$, best critical region adalah ${\prod }_{i=1}^n{X}_i\le k$ untuk suatu konstanta $k$

Karena bentuk critical region sama untuk semua ${\theta }_1<6$ (hanya konstanta $k$ yang berbeda, tergantung pada significance level), maka test yang menolak $H_0$ ketika ${\prod }_{i=1}^n{X}_i\le k$ adalah uniformly most powerful test.

$\therefore$ Statistik cukup untuk $\theta$ adalah $Y_1={\prod }_{i=1}^n{X}_i$. UMPT untuk $H_0:\theta =6$ melawan $H_1:\theta <6$ yang berkaitan dengan $Y_1$ adalah menolak $H_0$ ketika ${\prod }_{i=1}^n{X}_i\le k$ untuk suatu konstanta $k$.

9.28

Let $X_1,X_2,\dots ,X_n$ be a random sample from the normal distribution $N(\theta ,1)$. Show that the likelihood ratio principle for testing $H_0:\theta ={\theta }^{\prime }$, where ${\theta }^{\prime }$ is specified, against $H_1:\theta \ne {\theta }^{\prime }$ leads to the inequality $|\bar{x}-{\theta }^{\prime }|\ge c$. Is this a uniformly most powerful test of $H_0$ against $H_1$?

Jawab

Diketahui $X_i\sim N(\theta ,1)$. Fungsi likelihood untuk sampel adalah:

$$\begin{array}{rl}L(\theta )& =\prod _{i=1}^n\frac{1}{\sqrt{2\pi }}\exp \left(-\frac{(x_i-\theta {)}^2}{2}\right)\\ & =(2\pi {)}^{-n/2}\exp \left(-\frac{1}{2}\sum _{i=1}^n(x_i-\theta {)}^2\right)\\ & \\ \frac{\partial \ln L(\theta )}{\partial \theta }& =\frac{\partial }{\partial \theta }\left\{\ln \left[(2\pi {)}^{-n/2}\exp \left(-\frac{1}{2}\sum _{i=1}^n(x_i-\theta {)}^2\right)\right]\right\}\\ & =\frac{\partial }{\partial \theta }\left\{\ln \left[(2\pi {)}^{-n/2}\right]+\ln \left[\exp \left(-\frac{1}{2}\sum _{i=1}^n(x_i-\theta {)}^2\right)\right]\right\}\\ & =\frac{\partial }{\partial \theta }\left\{-\frac{1}{2}\sum _{i=1}^n(x_i-\theta {)}^2\right\}\\ 0& =\sum _{i=1}^n(x_i-\theta )\\ 0& =\sum _{i=1}^n{x}_i-n\theta \\ \theta & =\bar{x}\end{array}$$

Sehingga $\hat{\theta }=\bar{x}$ dan: $L(\hat{\theta })=L(\bar{x})=(2\pi {)}^{-n/2}\exp \left(-\frac{1}{2}{\sum }_{i=1}^n(x_i-\bar{x}{)}^2\right)$

Likelihood ratio adalah:

$$\begin{array}{rl}\lambda & =\frac{L({\theta }^{\prime })}{L(\hat{\theta })}\\ & =\frac{\exp \left(-\frac{1}{2}{\sum }_{i=1}^n(x_i-{\theta }^{\prime }{)}^2\right)}{\exp \left(-\frac{1}{2}{\sum }_{i=1}^n(x_i-\bar{x}{)}^2\right)}\\ & =\exp \left(-\frac{1}{2}\left[\sum _{i=1}^n(x_i-{\theta }^{\prime }{)}^2-\sum _{i=1}^n(x_i-\bar{x}{)}^2\right]\right)\end{array}$$

Perhatikan bahwa:

$$\begin{array}{rl}\sum _{i=1}^n(x_i-{\theta }^{\prime }{)}^2-\sum _{i=1}^n(x_i-\bar{x}{)}^2& =\sum _{i=1}^n[(x_i-{\theta }^{\prime }{)}^2-(x_i-\bar{x}{)}^2]\\ & =\sum _{i=1}^n[x_i^2-2x_i{\theta }^{\prime }+({\theta }^{\prime }{)}^2-x_i^2+2x_i\bar{x}-{\bar{x}}^2]\\ & =\sum _{i=1}^n[2x_i(\bar{x}-{\theta }^{\prime })+({\theta }^{\prime }{)}^2-{\bar{x}}^2]\\ & =2(\bar{x}-{\theta }^{\prime })\sum _{i=1}^n{x}_i+n[({\theta }^{\prime }{)}^2-{\bar{x}}^2]\\ & =2n(\bar{x}-{\theta }^{\prime }{)}^2+n[({\theta }^{\prime }{)}^2-{\bar{x}}^2]\\ & =n(\bar{x}-{\theta }^{\prime }{)}^2\end{array}$$

Sehingga: $\lambda =\exp \left(-\frac{n(\bar{x}-{\theta }^{\prime }{)}^2}{2}\right)$

Misalkan $k$ konstanta positif. Dapat diperoleh

$$\begin{array}{rl}\exp \left(-\frac{n(\bar{x}-{\theta }^{\prime }{)}^2}{2}\right)& \le k\\ -\frac{n(\bar{x}-{\theta }^{\prime }{)}^2}{2}& \le \ln k\\ (\bar{x}-{\theta }^{\prime }{)}^2& \ge -\frac{2\ln k}{n}\\ |\bar{x}-{\theta }^{\prime }|& \ge c\end{array}$$

dimana $c=\sqrt{-\frac{2\ln k}{n}}$.

Test ini bukan UMPT, karena UMPT didefiniskan untuk $H_1$ composite hypothesis, sementara $H_1$ yang diberikan adalah simple hypothesis.

$\therefore$ Diperoleh $|\bar{x}-{\theta }^{\prime }|\ge c$, tetapi test ini bukan UMPT karena $H_1$ adalah composite hypothesis.

9.47

Consider a random sample $X_1,X_2,\dots ,X_n$ from a distribution with pdf $f(x;\theta )=\theta (1-x{)}^{\theta -1},0<x<1$, zero elsewhere, where $\theta >0$.

1. Find the form of the uniformly most powerful test of $H_0:\theta =1$ against $H_1:\theta >1$.
2. What is the likelihood ratio for $\lambda$ for testing $H_0:\theta =1$ against $H_1:\theta \ne 1$?

9.47.1

Fungsi likelihood untuk sampel adalah:

$$\begin{array}{rl}L(\theta )& =\prod _{i=1}^n\theta (1-x_i{)}^{\theta -1}\\ & ={\theta }^n\prod _{i=1}^n(1-x_i{)}^{\theta -1}\\ & ={\theta }^n{\left[\prod _{i=1}^n(1-x_i)\right]}^{\theta -1}\end{array}$$

Misalkan ${\theta }_1>1$. Rasio likelihood adalah:

$$\begin{array}{rl}\frac{L({\theta }_1)}{L(1)}& =\frac{{\theta }_1^n{\left[{\prod }_{i=1}^n(1-x_i)\right]}^{{\theta }_1-1}}{1^n{\left[{\prod }_{i=1}^n(1-x_i)\right]}^{1-1}}\\ & ={\theta }_1^n{\left[\prod _{i=1}^n(1-x_i)\right]}^{{\theta }_1-1}\\ & \\ \ln \left(\frac{L({\theta }_1)}{L(1)}\right)& =n\ln ({\theta }_1)+({\theta }_1-1)\ln \left[\prod _{i=1}^n(1-x_i)\right]\\ & =n\ln ({\theta }_1)+({\theta }_1-1)\sum _{i=1}^n\ln (1-x_i)\end{array}$$

Karena ${\theta }_1>1$, maka ${\theta }_1-1>0$. Untuk $0<x_i<1$, kita memiliki $0<1-x_i<1$ sehingga $\ln (1-x_i)<0$.

Rasio likelihood meningkat ketika ${\sum }_{i=1}^n\ln (1-x_i)$ meningkat (menjadi kurang negatif), yang terjadi ketika ${\prod }_{i=1}^n(1-x_i)$ meningkat.

Oleh karena itu, untuk setiap ${\theta }_1>1$, best critical region adalah: ${\prod }_{i=1}^n(1-X_i)\ge k⟺{\sum }_{i=1}^n\ln (1-X_i)\ge k^{\prime },\forall {\theta }_1>1$ untuk suatu konstanta $k^{\prime }$.

$\therefore$ UMPT untuk $H_0:\theta =1$ versus $H_1:\theta >1$ adalah:

$\boxed{\text{Tolak }{H}_0\text{ jika }\sum _{i=1}^n\ln (1-X_i)\ge k^{\prime }}$

9.47.2

Untuk $H_0:\theta =1$:

$$\begin{array}{rl}L(1)& =\prod _{i=1}^n(1-x_i{)}^{1-1}=1\\ \ln L(\theta )& =n\ln (\theta )+(\theta -1)\sum _{i=1}^n\ln (1-x_i)\\ \frac{\partial \ln L}{\partial \theta }& =\frac{n}{\theta }+\sum _{i=1}^n\ln (1-x_i)=0\end{array}$$

Sehingga: $\hat{\theta }=-\frac{n}{{\sum }_{i=1}^n\ln (1-x_i)}$

Likelihood maksimum:

$$\begin{array}{rl}L(\hat{\theta })& ={\hat{\theta }}^n{\left[\prod _{i=1}^n(1-x_i)\right]}^{\hat{\theta }-1}\\ & ={\left(-\frac{n}{{\sum }_{i=1}^n\ln (1-x_i)}\right)}^n{\left[\prod _{i=1}^n(1-x_i)\right]}^{-\frac{n}{{\sum }_{i=1}^n\ln (1-x_i)}-1}\end{array}$$

Likelihood ratio adalah: $\boxed{\lambda =\frac{L(1)}{L(\hat{\theta })}=\frac{1}{L(\hat{\theta })}=\frac{{\left({\sum }_{i=1}^n\ln (1-x_i)\right)}^n}{(-n{)}^n}{\left[\prod _{i=1}^n(1-x_i)\right]}^{\frac{n}{{\sum }_{i=1}^n\ln (1-x_i)}+1}}$

Footnotes

1. Normal distribution ↩