Tugas Rangkuman 1

Soal

Soal no 1

Misalkan $X_1,X_2,X_3$ menyatakan sampel random berukuran 3 dari suatu distribusi bertipe kontinu dengan p.d.f. $f(x)$ yang positif pada $a<x<b$ dan sama dengan nol untuk yang lain, maka p.d.f. bersama dari $X_1,X_2,X_3$ adalah

$$f(x_1,x_2,x_3)=\{\begin{array}{ll}f(x_1)f(x_2)f(x_3),& a<x_i<b,i=1,2,3\\ 0,& (x_1,x_2,x_3)\text{ yang lain}\end{array}$$

Misalkan

$$\begin{array}{rl}{Y}_1& =\min (X_1,X_2,X_3),\\ Y_2& =\mathrm{mid}(X_1,X_2,X_3),\\ Y_3& =\max (X_1,X_2,X_3),\end{array}$$

Tentukan dicari p.d.f. bersama dari $Y_1,Y_2,Y_3$

Soal no 2

Misalkan $X_1,X_2,\dots ,X_n$ menyatakan sampel random berukuran $n$ dari suatu distribusi bertipe kontinu dengan p.d.f. $f(x)$ yang positif pada $a<x<b$ dan sama dengan nol untuk $x$ yang lain.

Misalkan $Y_1=\min (X_1,X_2,\dots ,X_n),Y_2=X_i$ berikutnya sesuai dengan urutan besarnya, $\dots$, dan $Y_n=\max (X_1,X_2,\dots ,X_n)$, maka $Y_i=i=1,2,\dots ,n$ disebut statistik terurut ke-$i$ dari sampel random $X_1,X_2,\dots ,X_n$ dan p.d.f. bersama dari $Y_1,Y_2,\dots ,Y_n$ adalah

$$g(y_1,y_2,\dots ,y_n)=\{\begin{array}{ll}(n!)f(y_1)f(y_2)\dots f(y_n),& a<y_1<y_2<\cdots <y_n<b\\ 0,& (y_1,y_2,\dots ,y_n)\text{ yang lain}\end{array}$$

• a. Tentukan pdf dari $Y_1$ dan $Y_n$
• b. Tentukan pdf bersama dari $Y_2$ dan $Y_4$

Soal no 3

Jelaskan secara lengkap mengenai pertidaksamaan Chebyshev dan berikan 2 contoh.

Jawab

Jawab no 1

Use Theorem 4.4.1 Joint pdf of order statistics

$$g(y_1,\dots ,y_n)=\{\begin{array}{ll}n!f(y_1)\dots f(y_n)& a<y_1<\cdots <y_n<b\\ 0& \text{elsewhere}\end{array}$$

Jawab no 2

a. Use Marginal pdf and of order statistics

$$\begin{array}{rl}{g}_1\left(y_1\right)& =\{\begin{array}{ll}n{\left[1-F\left(y_1\right)\right]}^{n-1}f\left(y_1\right),& a<y_1<b\\ 0,& \text{elsewhere}\end{array}\\ g_n\left(y_n\right)& =\{\begin{array}{ll}n{\left[F\left(y_n\right)\right]}^{n-1}f\left(y_n\right),& a<y_n<b\\ 0,& \text{elsewhere}\end{array}\end{array}$$

b. Use Marginal pdf and of order statistics

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

Jawab no 3

Explain using:

Theorem

Let:

• $X$ Random Variable
• ${\sigma }^2\in \mathbb{R}$ Variance of $X$
• $\mu =E(X)$ (by Existence of Lower Order Moments, ${\sigma }^2\in \mathbb{R}$ implies that $E(X)$ exists)

Then, for every $k>0$

$$P(|X-\mu |\ge k\sigma )\le \frac{1}{k^2}$$

Or equivalently,

$$P(|X-\mu |<k\sigma )\ge 1-\frac{1}{k^2}$$

Example

If $X$ is a random variable such that $E(X)=3$ and $E(X^2)=13$, use Chebyshev’s inequality to determine a lower bound for the probability $P(-2<X<8)$

We have,

• $\mu =E(X)=3$
• ${\sigma }^2=E(X^2)-[E(X){]}^2=13-3^2=4$
• $\sigma =\sqrt{{\sigma }^2}=\sqrt{4}=2$

Then

$$\begin{array}{rl}P(-2<X<8)& =P(-2-3<X-3<8-3)\\ & =P(-5<X-3<5)\\ & =P(|X-3|<5)\end{array}$$

By Chebyshev’s inequality, the form $P(|X-3|<5)$ has to satisfy $P(|X-\mu |<k\sigma )$. Thus

$$\begin{array}{rl}k\sigma & =5\\ k\cdot 2& =5\\ k& =\frac{5}{2}\end{array}$$

As a result, the lower bound for $P(-2<X<8)$ is:

$$\begin{array}{rl}P(-2<X<8)& =P(|X-3|<5)\\ & =1-\frac{1}{k^2}\\ & =1-\frac{1}{{\left(\frac{5}{2}\right)}^2}\\ & =1-\frac{4}{25}\\ & =\boxed{0.84}\end{array}$$Link to original

Example 1

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 $1-\frac{Y_n}{n}\stackrel{P}{\to }1-p$

Link to original

Example 2 Use proving Theorem 5.1.1 Weak law of large numbers