Chapter 5 Exercises

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 $1$.