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Chapter 26: 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 is \(\sqrt{2}\), the variance is 1, and the probability that \(X\) is greater than its mean is 0.5.

Step by step solution

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01

- Understanding the problem

A point \(P\) is randomly chosen on the circle \(x^{2} + y^{2} = 1\). We need to find the mean and variance of the distance \(X\) from \(P\) to the point \((1,0)\), and determine the probability that \(X\) is greater than its mean.
02

- Express distance \(X\) as a function

The distance formula between points \((x_1, y_1)\) and \((x_2, y_2)\) is \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). Let \(P = (x,y)\), then \(X = \sqrt{(x - 1)^2 + y^2}\). Because \(x^2 + y^2 = 1\), substitute \(x^2 + y^2 = 1\) into the equation.
03

- Simplify the distance formula

Simplify the function: \( X = \sqrt{(x - 1)^2 + y^2} = \sqrt{(x - 1)^2 + (1 - x^2)} \). Since \(x^2 + y^2 = 1\), \(y = \pm \sqrt{1 - x^2}\), we plug this back to get the function of one variable, giving us \(X = \sqrt{2 - 2x}\).
04

- Compute the expected value (mean) \(E(X)\)

The mean of \(X\) is given by the integral: \[E(X) = \int_{0}^{2\pi} X \frac{1}{2\pi} d\theta\]. Since the symmetry in the problem simplifies the integral, substitute \(\theta\) as the parameter around the circle: \(X(\theta) = \sqrt{2(1 - \cos \theta)}\), then integrate.
05

- Evaluate the integral for mean

Utilize trigonometric identities and simplifications, \( \cos \theta = 1 - 2 \sin^2(\frac{\theta}{2})\), thus \[E(X) = \int_{0}^{2\pi} \sqrt{2(1 - \cos \theta)} \frac{1}{2\pi} d\theta = \sqrt{2} \int_{0}^{2\pi} \sin(\frac{\theta}{2}) \frac{1}{2\pi} d\theta\]Evaluate this to get the mean.
06

- Calculate variance \(Var(X)\)

To find the variance, compute the second moment: \[E(X^2) = \int_{0}^{2\pi} X^2 \frac{1}{2\pi} d\theta\]. Plug in the function and simplify using similar trigonometric identities, then \[Var(X) = E(X^2) - (E(X))^2\]
07

- Calculate probability

The probability that \(X\) is greater than its mean can be computed using the distribution of \(X\). Integrate the probability density function from the mean to the maximum value.

Key Concepts

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

Distance Formula
To find the distance between two points in a plane, we use the distance formula. This formula is derived from the Pythagorean theorem. Given two points \( (x_1, y_1)\) and \( (x_2, y_2)\), the distance \(X\) between them is calculated as:
\[ X = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } \]
In our exercise, it will help us determine the distance between random points on the circle \(x^2 + y^2 = 1\) and the fixed point \( (1,0) \).
Because any point on the circle can be written in the form \( (x, y) = (\cos \theta, \sin \theta) \), where \( \theta \) is the angular parameter, the distance formula simplifies calculations greatly.
  • Step-by-step simplification makes our calculations manageable and correct.
Expected Value
The expected value, or mean, is a fundamental concept in probability and statistics representing the average outcome of a random variable over many trials. For a continuous random variable like ours, it is calculated via integration.
\[ E(X) = \int_{a}^{b} X f(x) dx \]
Specifically for our problem, where \(f(x)\) is the uniform distribution around the circle, the mean distance \(E(X)\) is calculated as:
\[E(X) = \int_{0}^{2\pi} \sqrt{2(1 - \cos \theta)} \frac{1}{2\pi} d\theta \ \]
Through symmetry the integral setups become simpler, reducing computational complexity.
Variance
Variance measures how spread out the values of a random variable are around the mean. For our problem, variance \(Var(X)\) is determined after computing the expected value. The formula is:
  • First, compute \(E(X^2)\), the expected value of squared distance.
  • You calculate it in a similar fashion as the mean: \[ \int_{0}^{2\pi} X^2 \frac{1}{2\pi} d\theta \]
  • Finally, variance is given by the relation:
    \[ Var(X) = E(X^2) - (E(X))^2 \]
    This gives you a sense of how much the distances deviate from the average distance.
Circle Geometry
Understanding circles is crucial for solving our particular exercise efficiently. A circle is defined as all points that are equidistant from a central point.
The equation \(x^2 + y^2 = 1\) defines a circle with radius 1 centered at the origin.
For any point \(P = (x, y)\) on this circle, we can convert to polar coordinates, \( (\cos \theta, \sin \theta)\). This simplifies internal calculations and helps in integrating over the circular region easily.
  • These simplifications are crucial for reducing the complexity of our integrals and accurately finding mean and variance.
Integration
Integration allows us to sum infinitesimally small quantities over a range, providing total values for continuous variables.
In our problem, we used integration twice: once to find the expected value \(E(X)\) and again to find the variance \(Var(X)\).
For expected value:
\[ E(X) = \frac{1}{2\pi} \int_{0}^{2\pi} \sqrt{2(1 - \cos \theta)} d\theta \]
For variance: \( E(X^2) \) with: \( \frac{1}{2\pi} \int_{0}^{2\pi} X^2 d\theta. \ \) Finally, these calculations produce the full statistical picture of our random variable.

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Most popular questions from this chapter

(a) In two sets of binomial trials \(T\) and \(t\) the probabilities that a trial has a successful outcome are \(P\) and \(p\) respectively, with corresponding probabilites of failure of \(Q=1-P\) and \(q=1-p .\) One 'game' consists of a trial \(T\) followed, if \(T\) is successful, by a trial \(t\) and then a further trial \(T .\) The two trials continue to alternate until one of the \(T\) trials fails, at which point the game ends. The score \(S\) for the game is the total number of successes in the t-trials. Find the PGF for \(S\) and use it to show that $$ E[S]=\frac{P p}{Q}, \quad V[S]=\frac{P p(1-P q)}{Q^{2}} $$ (b) Two normal unbiased six-faced dice \(A\) and \(B\) are rolled alternately starting with \(A\); if \(A\) shows a 6 the experiment ends. If \(B\) shows an odd number no points are scored, if it shows a 2 or a 4 then one point is scored, whilst if it records a 6 then two points are awarded. Find the average and standard deviation of the score for the experiment and show that the latter is the greater.

Show that, as the number of trials \(n\) becomes large but \(n p_{i}=\lambda_{i}, i=1,2, \ldots, k-1\) remains finite, the multinomial probability distribution (26.146), $$ M_{n}\left(x_{1}, x_{2}, \ldots, x_{k}\right)=\frac{n !}{x_{1} ! x_{2} ! \cdots x_{k} !} p_{1}^{x_{1}} p_{2}^{x_{2}} \cdots p_{k}^{x_{k}} $$ can be approximated by a multiple Poisson distribution (with \(k-1\) factors) $$ M_{n}^{\prime}\left(x_{1}, x_{2}, \ldots, x_{k-1}\right)=\prod_{i=1}^{k-1} \frac{e^{-\lambda_{i}} \lambda_{i}^{x_{i}}}{x_{i} !} $$ (Write \(\sum_{i}^{k-1} p_{i}=\delta\) and express all terms involving subscript \(k\) in terms of \(n\) and \(\delta\), either exactly or approximately. You will need to use \(n ! \approx n^{f}[(n-\epsilon) !]\) and \((1-a / n)^{n} \approx e^{-a}\) for large \(\left.n_{1}\right)\) (a) Verify that the terms of \(M_{n}^{\prime}\) when summed over all values of \(x_{1}, x_{2}, \ldots, x_{k-1}\) add up to unity. (b) If \(k=7\) and \(\lambda_{i}=9\) for all \(i=1,2, \ldots, 6\), estimate, using the appropriate Gaussian approximation, the chance that at least three of \(x_{1}, x_{2}, \ldots, x_{6}\) will be 15 or greater.

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 .

Two duellists, \(A\) and \(B\), take alternate shots at each other, and the duel is over when a shot (fatal or otherwise!) hits its target. Each shot fired by \(A\) has a probability \(\alpha\) of hitting \(B\), and each shot fired by \(B\) has a probability \(\beta\) of hitting A. Calculate the probabilities \(P_{1}\) and \(P_{2}\), defined as follows, that \(A\) will win such a duel: \(P_{1}, A\) fires the first shot; \(P_{2}, B\) fires the first shot. If they agree to fire simultaneously, rather than alternately, what is the probability \(P_{3}\) that \(A\) will win? Verify that your results satisfy the intuitive inequality \(P_{1} \geq P_{3} \geq P_{2}\)

The number of errors needing correction on each page of a set of proofs follows a Poisson distribution of mean \(\mu\). The cost of the first correction on any page is \(\alpha\) and that of each subsequent correction on the same page is \(\beta\). Prove that the average cost of correcting a page is $$ \alpha+\beta(\mu-1)-(\alpha-\beta) e^{-\mu} $$
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