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 XA,XB,XCX_A,X_B,X_CXA,XB,XC and XA,XB,XC∼iidU[−π,π]X_A,X_B,X_C \sim_{iid} U[-\pi,\pi]XA,XB,XC∼iidU[−π,π]. The paths from A to B can be counterclockwise or clockwise and the distance is ∣XA−XB∣|X_A - X_B|∣XA−XB∣ or 2π−∣XA−XB∣2\pi - |X_A - X_B|2π−∣XA−XB∣. Denote Y=∣XA−XB∣Y = |X_A - X_B|Y=∣XA−XB∣. The minimal distribution is given by
d1=min⁡(Y,2π−Y)d_1 = \min(Y,2\pi - Y)d1=min(Y,2π−Y)

Compute the distribution of XA−XBX_A - X_BXA−XB,
P(XA−XB≤x)=P(XA≤XB+x)=∫−ππdxB∫−πxB+x14π2dxA=x+π2πP(X_A - X_B \le x) = P(X_A \le X_B + x) = \int_{-\pi}^{\pi}dx_B \int_{-\pi}^{x_B+x} \frac{1}{4\pi^2}dx_A = \frac{x+\pi}{2\pi}P(XA−XB≤x)=P(XA≤XB+x)=∫−ππdxB∫−πxB+x4π21dxA=2πx+π

So XA−XB∼U[−π,π]⇒Y∼U[0,π]X_A - X_B \sim U[-\pi,\pi] \Rightarrow Y \sim U[0,\pi]XA−XB∼U[−π,π]⇒Y∼U[0,π],
P(d≤x)=1−P(min⁡(Y,2π−Y)>x)=1−P(Y>x)P(2π−Y>x)=1−π−xπ=xπP(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}P(d≤x)=1−P(min(Y,2π−Y)>x)=1−P(Y>x)P(2π−Y>x)=1−ππ−x=πx

Part b
To make the center of the circle included in the triangle ABC, given minimal distance of AB is xxx, C is restricted on the arc centrosymmetric with arc AB about the center. So the probability of center being covered is
x2π\frac{x}{2\pi}2πx

Part c
Denote the event center being covered as SSS, and we have
P(S∣d=x)=x2πP(S)=E[P(S∣d)]=E[X2π]=1/4P(S|d=x) = \frac{x}{2\pi} \\ P(S) = E[P(S|d)] = E[\frac{X}{2\pi}] = 1/4P(S∣d=x)=2πxP(S)=E[P(S∣d)]=E[2πX]=1/4

Notice
If XA,XB,XCX_A,X_B,X_CXA,XB,XC and XA,XB,XC∼iidU[0,2π]X_A,X_B,X_C \sim_{iid} U[0,2\pi]XA,XB,XC∼iidU[0,2π], compute the distribution of XA−XBX_A - X_BXA−XB,
P(XA−XB≤x)=P(XA≤XB+x)=∫02πdxB∫0xB+x14π2dxA=x+π2πP(X_A - X_B \le x) = P(X_A \le X_B + x) = \int_{0}^{2\pi}dx_B \int_{0}^{x_B+x} \frac{1}{4\pi^2}dx_A = \frac{x+\pi}{2\pi}P(XA−XB≤x)=P(XA≤XB+x)=∫02πdxB∫0xB+x4π21dxA=2πx+π

We can get the same answer.

第三题

Let X1=1X_1 = 1X1=1 denote the event that the first flip lands Head while X1=0X_1 = 0X1=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(X1=0)=1−pP(X_1 = 0) = 1-pP(X1=0)=1−p

Part b
Conditioning on the the first flip being a head, the next k−1k-1k−1 flips are head and the last flip is tail, so
P(X−1=k∣X1=1)=pk−1(1−p)P(X-1=k|X_1 = 1) = p^{k-1}(1-p)P(X−1=k∣X1=1)=pk−1(1−p)

So X−1∣X1=1∼Geo(1−p)X-1|X_1=1 \sim Geo(1-p)X−1∣X1=1∼Geo(1−p)

Part c
Similarly, P(X−1=k∣X1=0)=(1−p)k−1pP(X-1=k|X_1 = 0) = (1-p)^{k-1}pP(X−1=k∣X1=0)=(1−p)k−1p, X−1∣X1=0∼Geo(p)X-1|X_1=0\sim Geo(p)X−1∣X1=0∼Geo(p). So
E[X−1∣X1=1]=1p,E[X−1∣X1=0]=11−pE[X]=E[X−1]+1=1+1−pp+p1−pE[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}E[X−1∣X1=1]=p1, E[X−1∣X1=0]=1−p1E[X]=E[X−1]+1=1+p1−p+1−pp

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UA MATH564 概率论 QE练习题6

UA MATH564 概率论 QE练习题6

第二题

第三题

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