IS Math-2019-03

Source: Problem 3 Date: 2024-07-09 Topic: Math-Probability Theory-Geometric Probability

具体题目

Consider a triangle $ABC$ on a plane with vertices $A(1,0)$, $B(0,1)$, and $C(-1,-1)$. Let $\ell$ be a randomly chosen half-line originating from $(0,0)$, defined as:

$$\ell =\{(r\cos \Theta ,r\sin \Theta )\mid r\ge 0\}$$

where $\Theta$ is a uniformly distributed random variable on $[0,2\pi )$. Let $Q$ be the intersection point of $\ell$ and the edges of triangle $ABC$, with coordinates $(X,Y)$. Answer the following questions:

1. Find the probability that $Q$ is located on segment $AB$.

2. Prove that $\mathbb{E}[X\mid Q\text{ on }AB]=1/2$, given that $ABC$ is symmetric about $y=x$.

3. Derive the probability density function of $X$ given $Q$ is on $BC$, using the change-of-variables formula:

$$f(x)=g(h(x))\left|\frac{\mathrm{d}h}{\mathrm{d}x}(x)\right|$$

where $f$ and $g$ are the PDFs of $X$ and $\Theta$ respectively, and $\Theta =h(X)$.

1. Calculate $\alpha =\mathbb{E}[X\mid Q\text{ on }BC]$ using the result from (3).

2. Determine the unconditional expectation $\mu =\mathbb{E}[X]$.

正确解答

1. Probability of $Q$ on Segment $AB$

For $Q$ to lie on $AB$, $\Theta$ must be between the angles of vectors $\mathbf{OA}$ and $\mathbf{OB}$.

$\angle AOX=0$ (as $\mathbf{OA}=(1,0)$)

$\angle BOX=\pi /2$ (as $\mathbf{OB}=(0,1)$)

Given $\Theta \sim U[0,2\pi )$, the probability is:

$$P(Q\text{ on }AB)=\frac{\pi /2-0}{2\pi }=\frac{1}{4}$$

2. Expectation of $X$ Given $Q$ on $AB$

The line $y=x$ bisects $ABC$ symmetrically. Thus, the PDF of $X$ given $Q$ on $AB$ is symmetric about $x=1/2$:

$$P(x\mid Q\text{ on }AB)=P(1-x\mid Q\text{ on }AB)$$

This implies:

$$\mathbb{E}[X\mid Q\text{ on }AB]=\mathbb{E}[1-X\mid Q\text{ on }AB]$$ $$\mathbb{E}[X\mid Q\text{ on }AB]=1-\mathbb{E}[X\mid Q\text{ on }AB]$$

Therefore:

$$\mathbb{E}[X\mid Q\text{ on }AB]=\frac{1}{2}$$

3. PDF of $X$ Given $Q$ on $BC$

• Step 1: Equation of segment $BC$ $B(0,1)$, $C(-1,-1)$ Slope of $BC$: $\frac{-1-1}{-1-0}=2$ Equation: $y=2x+1$ for $x\in [-1,0]$
• Step 2: Parameterize $Q$ $Q(X,Y)=(x,2x+1)$
• Step 3: Express $\Theta$ in terms of $X$ $\Theta =\arctan (\frac{2x+1}{x})=\arctan \left(2+\frac{1}{x}\right)$
• Step 4: Find $g(\Theta )$ $\Theta \in [\frac{\pi }{2},\frac{5\pi }{4}]$ when $Q$ is on $BC$ $g(\Theta )=\frac{1}{\frac{5\pi }{4}-\frac{\pi }{2}}=\frac{4}{3\pi }$
• Step 5: Apply the change-of-variables formula:

$$\frac{\mathrm{d}h}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}\arctan \left(2+\frac{1}{x}\right)=\frac{-1}{x^2\left(1+{\left(2+\frac{1}{x}\right)}^2\right)}$$

• Step 6: Substituting into the formula:

$$f(x)=\frac{4}{3\pi }\cdot \frac{1}{x^2\left(1+{\left(2+\frac{1}{x}\right)}^2\right)}=\frac{4}{3\pi }\cdot \frac{1}{1+4x+5x^2}$$

Therefore, the PDF of $X$ given $Q$ on $BC$ is:

$$f(x)=\frac{4}{3\pi }\cdot \frac{1}{1+4x+5x^2}$$

4. Expectation of $X$ Given $Q$ on $BC$

$$\alpha =\mathbb{E}[X\mid Q\text{ on }BC]={\int }_{-1}^{0}xf(x)\mathrm{d}x=\frac{4}{3\pi }{\int }_{-1}^0\frac{x}{1+4x+5x^2}\mathrm{d}x$$

Let $u=1+4x+5x^2$, then $\mathrm{d}u=(4+10x)\mathrm{d}x$

$$\alpha =\frac{4}{3\pi }\left(\frac{1}{10}{\int }_2^1\frac{\mathrm{d}u}{u}-\frac{2}{5}{\int }_{-1}^0\frac{1}{1+4x+5x^2}\mathrm{d}x\right)$$ $${\int }_2^1\frac{\mathrm{d}u}{u}=\ln |u|{|}_2^1=\ln \frac{1}{2}=-\ln 2$$

For the second integral, let $v=x+\frac{2}{5}$:

$${\int }_{-1}^0\frac{1}{1+4x+5x^2}\mathrm{d}x=\frac{1}{5}{\int }_{-\frac{3}{5}}^{\frac{2}{5}}\frac{1}{v^2+(\frac{1}{5}{)}^2}\mathrm{d}v=\arctan 5v{|}_{-\frac{3}{5}}^{\frac{2}{5}}=\arctan 2-\arctan (-3)$$ $$\arctan 2-\arctan (-3)=\arctan \left(\frac{2+3}{1-2\cdot 3}\right)+\pi =\arctan (-1)+\pi =-\frac{\pi }{4}+\pi =\frac{3\pi }{4}$$

Substituting back:

$$\alpha =\frac{4}{3\pi }\left(-\frac{\ln 2}{10}-\frac{3\pi }{10}\right)=-\frac{2\ln 2}{15\pi }-\frac{2}{5}$$

5. Unconditional Expectation of $X$

Using the law of total expectation:

$$\mu =\mathbb{E}[X]=P(Q\text{ on }AB)\mathbb{E}[X\mid Q\text{ on }AB]+P(Q\text{ on }BC)\mathbb{E}[X\mid Q\text{ on }BC]+P(Q\text{ on }CA)\mathbb{E}[X\mid Q\text{ on }CA]$$

By symmetry, $\mathbb{E}[Y\mid Q\text{ on }CA]=\mathbb{E}[X\mid Q\text{ on }BC]$

The line $CA$ can be parameterized as $x=2y+1$

$$\mathbb{E}[X\mid Q\text{ on }CA]=2\mathbb{E}[Y\mid Q\text{ on }CA]+1=2\mathbb{E}[X\mid Q\text{ on }BC]+1=2\alpha +1=-\frac{4\ln 2}{15\pi }+\frac{1}{5}$$

The probabilities for $BC$ and $CA$ are each $\frac{3\pi /4}{2\pi }=\frac{3}{8}$

Therefore:

$$\mu =\frac{1}{4}\cdot \frac{1}{2}+\frac{3}{8}\cdot \left(-\frac{2\ln 2}{15\pi }-\frac{2}{5}\right)+\frac{3}{8}\cdot \left(-\frac{4\ln 2}{15\pi }+\frac{1}{5}\right)=\frac{1}{20}-\frac{3\ln 2}{10\pi }$$

知识点

几何概率期望值概率密度函数换元积分法

Calculation Techniques

难点解题思路

This problem demonstrates the application of geometric probability and conditional expectation in a triangular setting. The key steps involve:

1. Utilizing the symmetry of the triangle to simplify calculations.
2. Applying the change-of-variables formula to derive probability density functions.
3. Using integration techniques to calculate expectations.
4. Leveraging the law of total expectation to combine conditional expectations.

解题技巧和信息

• 换元积分法 在求解 $X$ 在 $BC$ 上的条件概率密度函数时，我们使用了换元积分法。具体来说，我们将 $\Theta =\arctan (2+\frac{1}{x})$ 代入概率密度函数的变换公式中。
• 分部积分法 虽然在最终的解答中没有明确使用，但在处理 $\int xf(x)dx$ 形式的积分时，分部积分法是一个潜在的有用工具。
• 有理函数的积分 在计算 $\alpha =E[X|Q\text{ on }BC]$ 时，我们遇到了形如 $\int \frac{x}{1+4x+5x^2}dx$ 的积分。这是一个有理函数的积分，通常需要使用部分分式分解或替换等技巧。
• 三角函数的积分 在求解上述有理函数积分时，我们使用了替换 $u=1+4x+5x^2$，最终得到了 $\arctan$ 函数，这涉及到了三角函数的积分。

Techniques and Insights

1. Symmetry can greatly simplify probabilistic calculations in geometric settings.
2. The change-of-variables formula is crucial for deriving probability density functions in transformed spaces.
3. Breaking down complex integrals into manageable parts can facilitate their evaluation.
4. Understanding the relationships between different segments of a geometric figure can help in extending results from one part to another.

Key Terminology

• Geometric Probability
• Conditional Expectation
• Probability Density Function (PDF)
• Change-of-Variables Formula
• Law of Total Expectation
• Symmetry in Probability