UA MATH564 概率论 QE练习题6

• 第二题
• 第三题

这是2016年5月的2-3题。

第二题

Part a
Denote O as center of the circle. Randomly choose point Don the unit circle and represent points A,B,C using the center angle counterclockwise from OD to OA,OB,OC and denote as $X_A,X_B,X_C$ and $X_A,X_B,X_C{\sim }_{iid}U[-\pi ,\pi ]$. The paths from A to B can be counterclockwise or clockwise and the distance is $|X_A-X_B|$ or $2\pi -|X_A-X_B|$. Denote $Y=|X_A-X_B|$. The minimal distribution is given by
$$d_1=\min (Y,2\pi -Y)$$

Compute the distribution of $X_A-X_B$,
$$P(X_A-X_B\le x)=P(X_A\le X_B+x)={\int }_{-\pi }^{\pi }d{x}_B{\int }_{-\pi }^{x_B+x}\frac{1}{4{\pi }^2}d{x}_A=\frac{x+\pi }{2\pi }$$

So $X_A-X_B\sim U[-\pi ,\pi ]\Rightarrow Y\sim U[0,\pi ]$,
$$\begin{gathered} P(d\le x)=1-P(\min (Y,2\pi -Y)>x) \\ =1-P(Y>x)P(2\pi -Y>x)=1-\frac{\pi -x}{\pi }=\frac{x}{\pi } \end{gathered}$$

Part b
To make the center of the circle included in the triangle ABC, given minimal distance of AB is $x$, C is restricted on the arc centrosymmetric with arc AB about the center. So the probability of center being covered is
$$\frac{x}{2\pi }$$

Part c
Denote the event center being covered as $S$, and we have
$$\begin{gathered} P(S|d=x)=\frac{x}{2\pi } \\ P(S)=E[P(S|d)]=E[\frac{X}{2\pi }]=1/4 \end{gathered}$$

Notice
If $X_A,X_B,X_C$ and $X_A,X_B,X_C{\sim }_{iid}U[0,2\pi ]$, compute the distribution of $X_A-X_B$,
$$P(X_A-X_B\le x)=P(X_A\le X_B+x)={\int }_0^{2\pi }d{x}_B{\int }_0^{x_B+x}\frac{1}{4{\pi }^2}d{x}_A=\frac{x+\pi }{2\pi }$$

We can get the same answer.

第三题

Let $X_1=1$ denote the event that the first flip lands Head while $X_1=0$ denote the event that the first flip lands Tail.
Part a
Only if the first flip lands Tail, then last flip lands Head. So the probability of the event that the last flip lands Head is
$$P(X_1=0)=1-p$$

Part b
Conditioning on the the first flip being a head, the next $k-1$ flips are head and the last flip is tail, so
$$P(X-1=k|X_1=1)=p^{k-1}(1-p)$$

So $X-1|X_1=1\sim Geo(1-p)$

Part c
Similarly, $P(X-1=k|X_1=0)=(1-p{)}^{k-1}p$, $X-1|X_1=0\sim Geo(p)$. So
$$\begin{gathered} E[X-1|X_1=1]=\frac{1}{p},E[X-1|X_1=0]=\frac{1}{1-p} \\ E[X]=E[X-1]+1=1+\frac{1-p}{p}+\frac{p}{1-p} \end{gathered}$$

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