Question

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

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

- 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.

- 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.

- 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 \(X = \sqrt{2 - 2x}\).

- 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.

- 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.

- 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 \[Var(X) = E(X^2) - (E(X))^2\]

- 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

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.