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Chapter 30: Problem 23

A point \(P\) is chosen at random on the circle \(x^{2}+y^{2}=1\). The random variable \(X\) denotes the distance of \(P\) from \((1,0)\). Find the mean and variance of \(X\) and the probability that \(X\) is greater than its mean.

Short Answer

Expert verified
The mean of X is 1, the variance of X is \( \frac{1}{3} \), and the probability that X is greater than its mean is \( \frac{1}{2} \).

Step by step solution

01

- Understand the Point on the Circle

A random point P on the circle with equation \( x^2 + y^2 = 1 \) means that P has coordinates \( ( \text{cos} \theta, \text{sin} \theta ) \) where \( \theta \) is uniformly distributed over \( [0, 2\text{蟺}] \).
02

- Calculate the Distance X

The distance \( X \) from the point P to (1,0) can be expressed as: \[ X = \text{distance}(( \text{cos} \theta , \text{sin} \theta ), (1,0)) = \text{鈭殅(( \text{cos} \theta - 1)^2 + ( \text{sin} \theta)^2) = \text{鈭殅(1 - 2 \text{cos} \theta + \text{cos}^2 \theta + \text{sin}^2 \theta) = \text{鈭殅(2 - 2 \text{cos} \theta) = \text{鈭殅2 \text{鈭殅(1 - \text{cos} \theta) \]
03

- Use the Double-Angle Formula

Using the identity \( \text{cos} \theta = 1 - 2 \text{sin}^2 \frac{\theta}{2} \), we have: \[ X = \text{鈭殅2 \text{鈭殅(1 - (1 - 2 \text{sin}^2 \frac{\theta}{2})) = 2 \text{sin} \frac{\theta}{2} \]
04

- Find the Mean of X

For the random variable\( Y = \text{sin} \frac{\theta}{2} \), it is uniformly distributed in the range [0,1]. The mean of\( Y is \frac{1}{2} \). Since \( X = 2Y \), the mean of \( X is 2 \times \frac{1}{2} = 1. \)
05

- Calculate the Variance of X

The variance of\( Y is \text{Var}(Y) = \frac{1}{12} \). The variance of\( X = 2X is \text{Var}(X) = 4 \times \frac{1}{12} = \frac{1}{3} \).
06

- Probability that X is Greater than its Mean

Given \( X \) is uniformly distributed in the range [0,2], the mean of\( X is 1 \). Thus,\( P(X > 1) = P(\text{sin} \frac{\theta}{2} > \frac{1}{2}) = \text{the probability of} \frac{\theta}{2} > \frac{\text{蟺}}{6} \). This is\( \frac{2\text{蟺}-\text{蟺}}{2\text{蟺}} = \frac{1}{2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Circle Equation
When dealing with geometrical problems like the one in this exercise, understanding the circle equation is crucial. The circle equation given is \(x^2 + y^2 = 1\). This represents a circle with a radius of 1 centered at the origin (0,0). Any point \(P\) on this circle can be expressed in terms of an angle \(\theta\). Specifically, the coordinates of \(P\) are \((\cos\theta, \sin\theta)\), where \(\theta\) varies from 0 to \(2\pi\). This parameterization is key to solving the problem because it links the random variable to the geometric representation.
Uniform Distribution
The problem mentions that the angle \(\theta\) is uniformly distributed over \([0, 2\pi]\). In probability, a variable is uniformly distributed when every outcome in its range is equally likely. For \(\theta\), this means any angle between 0 and \(2\pi\) is just as probable as any other. This is important when finding probabilities and means. Given this uniform distribution, calculations involving \(\sin\frac{\theta}{2}\) become straightforward because the values are evenly spread out.
Variance Calculation
Variance measures how spread out a set of numbers is. In our case, we calculate the variance of \(X\). First, we find the variance of \(Y = \sin\frac{\theta}{2}\), which is uniformly distributed over \([0,1]\). The variance of a uniformly distributed variable from 0 to 1 is \(\frac{1}{12}\). Since \(X = 2Y\), the variance scales by the square of the constant factor. Therefore, the variance of \(X\) becomes: \[ \text{Var}(X) = 4 \times \text{Var}(Y) = 4 \times \frac{1}{12} = \frac{1}{3} \]. This result shows how much the random distance deviates from the mean.
Mean of a Random Variable
The mean, or expected value, provides a measure of the central tendency of a random variable. In this exercise, we find the mean of \(X\). We use \(Y = \sin\frac{\theta}{2}\), which is uniformly distributed on \([0,1]\). The mean of \(Y\) is \(\frac{1}{2}\). Since \(X = 2Y\), the mean of \(X\) becomes: \[ \text{E}(X) = 2 \times \text{E}(Y) = 2 \times \frac{1}{2} = 1 \]. This calculated mean is useful for further analysis, such as finding the probability that \(X\) exceeds its mean.

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Most popular questions from this chapter

By shading or numbering Venn diagrams, determine which of the following are valid relationships between events. For those that are, prove the relationship using de Morgan's laws. (a) \(\overline{(\bar{X} \cup Y)}=X \cap \bar{Y}\). (b) \(\bar{X} \cup \bar{Y}=\overline{(X \cup Y)}\). (c) \((X \cup Y) \cap Z=(X \cup Z) \cap Y\). (d) \(X \cup \underline{(Y \cap Z)}=(X \cup \underline{Y}) \cap Z\). (e) \(X \cup \overline{(Y \cap Z)}=(X \cup \bar{Y}) \cup \bar{Z}\)

A continuous random variable \(X\) has a probability density function \(f(x) ;\) the corresponding cumulative probability function is \(F(x)\). Show that the random variable \(Y=F(X)\) is uniformly distributed between 0 and 1 .

The variables \(X_{i}, i=1,2, \ldots, n\), are distributed as a multivariate Gaussian, with means \(\mu_{i}\) and a covariance matrix \(\mathrm{V} .\) If the \(X_{i}\) are required to satisfy the linear constraint \(\sum_{i=1}^{n} c_{i} X_{i}=0\), where the \(c_{i}\) are constants (and not all equal to zero), show that the variable $$ \chi_{n}^{2}=(\mathrm{x}-\mu)^{\mathrm{T}} \mathrm{V}^{-1}(\mathrm{x}-\mu) $$ follows a chi-squared distribution of order \(n-1\)

A tennis tournament is arranged on a straight knockout basis for \(2^{n}\) players, and for each round, except the final, opponents for those still in the competition are drawn at random. The quality of the field is so even that in any match it is equally likely that either player will win. Two of the players have surnames that begin with ' \(Q\) '. Find the probabilities that they play each other (a) in the final, (b) at some stage in the tournament.

A particle is confined to the one-dimensional space \(0 \leq x \leq a\), and classically it can be in any small interval \(d x\) with equal probability. However, quantum mechanics gives the result that the probability distribution is proportional to \(\sin ^{2}(n \pi x / a)\), where \(n\) is an integer. Find the variance in the particle's position in both the classical and quantum- mechanical pictures, and show that, although they differ, the latter tends to the former in the limit of large \(n\), in agreement with the correspondence principle of physics.
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