Probability HW5

tags: Probability

1

a

以下計算

$M_X(t)$

$$\begin{array}{rl}{M}_X(t)& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{xt}{f}_X(x)\mathrm{d}x\\ & ={\int }_{-\mathrm{\infty }}^{-1}{e}^{xt}(0)\mathrm{d}x+{\int }_{-1}^3{e}^{xt}\left(\frac{1}{4}\right)\mathrm{d}x+{\int }_3^{\mathrm{\infty }}{e}^{xt}(0)\mathrm{d}x\\ & =0+{\left[\frac{1}{4t}{e}^{xt}\right]}_{-1}^3+0\\ & =\frac{e^{3t}-e^{-t}}{4t}\end{array}$$

b

已知

${M_X}^{(k)}(0)=E[X^k]$ 。因此

• $E[X]={M_X}^{\prime }(0)$
• $\text{Var}[X]=E[X^2]-E[X]={M_X}^{″}(0)-{M_X}^{\prime }(0)$

以下計算

$E[X]$

$$\begin{array}{rl}E[X]& ={M_X}^{\prime }(0)\\ & =\underset{t\to 0}{lim}\frac{(3e^{3t}+e^{-t})(4t)-(e^{3t}-e^{-t})(4)}{(4t{)}^2}\\ & =\underset{t\to 0}{lim}\frac{t(3e^{3t}+e^{-t})-(e^{3t}-e^{-t})}{4t^2}\\ & \stackrel{\text{H}}{=}\underset{t\to 0}{lim}\frac{(3e^{3t}+e^{-t})+t(9e^{3t}-e^{-t})-(3e^{3t}+e^{-t})}{8t}\\ & =\underset{t\to 0}{lim}\frac{t(9e^{3t}-e^{-t})}{8t}\\ & \stackrel{\text{H}}{=}\underset{t\to 0}{lim}\frac{(9e^{3t}-e^{-t})+t(27e^{3t}+e^{-t})}{8}\\ & =1\end{array}$$

以下計算

$\text{Var}[X]$

$$\begin{array}{rl}\text{Var}[X]& ={M_X}^{″}(0)-{M_X}^{\prime }(0)\\ & =\underset{t\to 0}{lim}{\displaystyle \frac{(9t^2-6t+2){\mathrm{e}}^{4t}-t^2-2t-2}{4t^3{e}^t}}-1\\ & \stackrel{\text{H}}{=}\underset{t\to 0}{lim}\frac{(36t^2-6t+2)e^{4t}-2t-2}{4e^t(3t^2+t^3)}-1\\ & \stackrel{\text{H}}{=}\underset{t\to 0}{lim}\frac{(144t^2+48t+2)e^{4t}-2}{4e^t(6t+6t^2+t^3)}-1\\ & \stackrel{\text{H}}{=}\underset{t\to 0}{lim}\frac{(576t^2+488t+56)e^{4t}}{4e^t(6+18t+9t^2+t^3)}-1\\ & =\frac{4}{3}\end{array}$$

2

a

顯然這是

$n=7,p=0.75$ 的 Binomial distribution 。因此

$$Pr(X=x)=\{\begin{array}{ll}{\displaystyle (\genfrac{}{}{0ex}{}{7}{x}){\left(\frac{3}{4}\right)}^x{\left(\frac{1}{4}\right)}^{7-x},}& \text{if}x\in \{0,1,2,3,4,5,6,7\}\\ \\ 0,& \text{else}\end{array}$$

b

顯然這是 Geometric distribution 。解

$$\frac{p{e}^x}{1-(1-p)e^x}=\frac{e^x}{2-e^x}$$

得到

$p=0.5$ ，因此

$$Pr(X=x)=\{\begin{array}{ll}{2}^{-x},& \text{if}x\in \mathbb{N}\\ 0,& \text{else}\end{array}$$

c

顯然這是

$\lambda =3$ 的 Poisson distribution 。因此

$$Pr(X=x)=\{\begin{array}{ll}{\displaystyle \frac{3^x}{x!e^3}},& \text{if}x\in \mathbb{N}\text{or}x=0\\ \\ 0,& \text{else}\end{array}$$

3

a

已知

$$f_{X_1}(x)=f_{X_2}(x)=\{\begin{array}{ll}0.5,& \text{if}x\in [0,2]\\ 0,& \text{else}\end{array}$$

可以推導出

$$f_{X_1}(t-x)=\{\begin{array}{ll}0.5,& \text{if}x\in [t-2,t]\\ 0,& \text{else}\end{array}$$

因此

• 當
  $x<0$ 時
  $f_{X_1+X_2}(t)={\displaystyle {\int }_{-\mathrm{\infty }}^0{0}^2\mathrm{d}x=0}$
• 當
  $x\in [0,2]$ 時
  $f_{X_1+X_2}(t)={\displaystyle {\int }_0^t{0.5}^2\mathrm{d}x={\displaystyle \frac{t}{4}}}$
• 當
  $x\in [2,4]$ 時
  $f_{X_1+X_2}(t)={\displaystyle {\int }_t^4{0.5}^2\mathrm{d}x=1-{\displaystyle \frac{t}{4}}}$
• 當
  $x>4$ 時
  $f_{X_1+X_2}(t)={\displaystyle {\int }_4^{\mathrm{\infty }}{0}^2\mathrm{d}x=0}$

答案為

$$f_{X_1+X_2}(t)=\{\begin{array}{ll}{\displaystyle \frac{t}{4}},& \text{if}t\in [0,2]\\ 1-{\displaystyle \frac{t}{4}},& \text{if}t\in [2,4]\\ 0,& \text{else}\end{array}$$

b

已知

$Z=3Y_1+4Y_2=\mathcal{N}(3{\mu }_{Y_1}+4{\mu }_{Y_2},3^2{{\sigma }_{Y_1}}^2+4^2{{\sigma }_{Y_2}}^2)=\mathcal{N}(19,180)$ 。 因此

$$\begin{array}{rl}Pr(Z>20)& =1-\frac{1}{2}(1+\text{erf}\left(\frac{20-19}{\sqrt{180}\sqrt{2}}\right))\\ & =1-\frac{1}{2}(1+\text{erf}\left(\frac{1}{\sqrt{360}}\right))\\ & \approx 0.47029\end{array}$$

4

a

已知

$$\begin{array}{rl}Pr(X>\pi \mu )& =Pr(X-\mu >(\pi -1)\mu )\\ & \le Pr(|X-\mu |>(\pi -1)\mu )\\ & \le \frac{\mu }{((\pi -1)\mu {)}^2}\\ & =\frac{1}{(\pi -1{)}^2\mu }\end{array}$$

所以

$$Pr(X>\pi \mu )\le \frac{1}{(\pi -1{)}^2\mu }$$

b

令

$H(n)$ 表示執行

$n$ 次演算法正確的次數。根據 Hoeffding's inequality 若

$a\in [0,1]$ ，則有

$$Pr(H(n)\le (p-a)n)\le \exp (-2a^{2}n)$$

在此例中

$p=0.5+\delta$ ，並且若

$H(n)\le n/2$ 則預測失敗。解

$(p-a)n=n/2$ 得到

$a=\delta$ 。故預測失敗的機率為

$$Pr(H(n)\le \frac{n}{2})\le \exp (-2{\delta }^{2}n)$$

當

$n\ge (1/2){\delta }^{-2}\ln ({\epsilon }^{-1})$ 時，預測失敗的機率將為

$$\begin{array}{rl}Pr(H(n)\le \frac{n}{2})& \le \exp (-2{\delta }^{2}n)\\ & \le \exp (-2{\delta }^2\frac{1}{2{\delta }^2}\ln \left(\frac{1}{\epsilon }\right))\\ & =\epsilon \end{array}$$

故得知失敗機率至多為

$\epsilon$ ，亦即預測成功機率至少為

$1-\epsilon$ 。 Q.E.D 。

5

a

已知

$f_{X_i}(x)\le 1$ ，因此

$$\begin{array}{rl}E[\exp (-t{X}_i)]& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-tx}\cdot f_{X_i}(x)\mathrm{d}x\\ & \le {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-tx}\cdot 1\mathrm{d}x\\ & =\frac{1}{t}\end{array}$$

b

I don't know.