Question

A certain marksman never misses his target, which consists of a disc of unit radius with centre \(O .\) The probability that any given shot will hit the target within a distance \(t\) of \(O\) is \(t^{2}\) for \(0 \leq t \leq 1\). The marksman fires \(n\) independendent shots at the target, and the random variable \(Y\) is the radius of the smallest circle with centre \(O\) that encloses all the shots. Determine the PDF for \(Y\) and hence find the expected area of the circle. The shot that is furthest from \(O\) is now rejected and the corresponding circle determined for the remaining \(n-1\) shots. Show that its expected area is $$ \frac{n-1}{n+1} \pi $$

Short answer

The PDF of Y is f_Y(y) = 2n y^(2n-1) for 0 ≤ y ≤ 1. The expected area of the circle with n shots is π * n/(n+1). If the furthest shot is rejected, the expected area for the remaining n-1 shots is π(n-1)/(n+1).

Step-by-step solution

Define the Problem's Variables

Let the random variable Y represent the radius of the smallest circle that encloses all n shots. Each shot has a probability density function of hitting within a radius t given by t^2 for 0 ≤ t ≤ 1.

Understand Distribution of Y

Since Y is the radius of the smallest circle enclosing all shots, it's the maximum radius of any individual shot. Define Y = max(T_1, T_2, ..., T_n), where T_i are the radii of individual shots.

Compute Cumulative Distribution Function (CDF) of Y

To find the CDF of Y, P(Y ≤ y) = P(max(T_1, T_2, ..., T_n) ≤ y). Each shot's radius must be less than or equal to y, so P(T_i ≤ y) = y^2. Therefore, P(Y ≤ y) = (y^2)^n. This is valid for 0 ≤ y ≤ 1.

Derive the Probability Density Function (PDF)

The PDF of Y, f_Y(y), is obtained by differentiating the CDF with respect to y. So, f_Y(y) = d/dy [(y^2)^n] = 2n y^(2n-1) for 0 ≤ y ≤ 1.

Determine the Expected Area of the Circle

The expected value E[Y^2] represents the expected square of the radius, thus the expected area is πE[Y^2]. Using the PDF, calculate E[Y^2] = ∫_0^1 y^2 f_Y(y) dy = ∫_0^1 y^2 * 2n y^(2n-1) dy = 2n ∫_0^1 y^(2n+1) dy = 2n / (2n+2) = n/(n+1). Hence, the expected area is E[Area] = πE[Y^2] = π * n/(n+1).

Reject the Furthest Shot and Calculate the Expected Area

When the furthest shot is rejected, consider Y' to be the radius of the smallest circle that encloses the remaining n-1 shots. Repeating the steps for Y', we find E[(Y')^2] = (n-1)/(n+1). Thus, the expected area for the remaining shots is πE[(Y')^2] = π(n-1)/(n+1).

Key concepts

Cumulative Distribution Function

The Cumulative Distribution Function (CDF) tells us the probability that a random variable takes on a value less than or equal to a specific value. It's useful in understanding how the values of a variable are distributed.
In our exercise, the CDF is used to understand the radius of the smallest circle, denoted as Y, that encloses all the shots. When we say P(Y ≤ y), it represents the probability that the radius of the smallest circle is less than or equal to y.
Since every individual shot has a radius distributed as per the function t^2, the combined probability, considering n shots, is determined by raising the probability function to the power of n. In other words, P(Y ≤ y) = (y^2)^n for 0 ≤ y ≤ 1. This means that the more shots fired, the higher the power, leading to a richer understanding of the overall distribution.
By knowing the CDF, we can then proceed to derive other important properties of the random variable Y.

Expected Value

Expected value, also known as the mean, is a fundamental concept in probability theory. It represents the average or mean value that one can expect from a random variable, essentially giving a central value around which the values of the variable spread.
In this exercise, we want to find the expected area of the circle using the expected value of Y^2. To get there, we use the Probability Density Function (PDF) derived from the CDF. The formula for the expected value E[Y^2] involves integrating Y^2 times the PDF over the permissible range (0 to 1 in this case).
This gives us E[Y^2] = ∫_0^1 y^2 * 2n y^(2n-1) dy. When simplified, it results in E[Y^2] = n / (n+1). Therefore, the expected area is πE[Y^2], simplifying to π * n/(n+1).
Understanding the expected value helps us find the typical size of the smallest enclosing circle's area, making it crucial for solving the problem.

Independent Random Variables

Independent random variables are variables whose outcomes do not depend on each other. This means that the occurrence of one event does not influence the probability of the occurrence of another.
In our exercise, the n shots fired at the target are considered independent. Each shot's radius falls within a specific distribution independently of the others. This independence allows us to treat each shot separately when calculating probabilities and then combine the individual probabilities to find the overall distribution.
For instance, to find the PDF of the radius Y, we leverage the independence of each shot and compute the maximum radius, leading us to the expression (y^2)^n. The independence simplifies complex calculations since we do not need to account for interdependencies among the shots.
Recognizing that the shots are independent makes the problem tractable and ensures that our use of probability theory principles is valid.

Probability Distribution

A probability distribution tells us how probabilities are assigned to the values of a random variable. It can take the form of a CDF, PDF, or a discrete list of probabilities for certain outcomes.
In this exercise, the shots' radii follow a specific probability distribution defined by the function t^2 within the range 0 ≤ t ≤ 1. This forms the basis for finding the cumulative distribution and subsequently the probability density function.
By understanding the distribution, we can calculate critical aspects such as the CDF (P(Y ≤ y) = (y^2)^n) and then derive the PDF (f_Y(y) = 2n y^(2n-1)). This detailed understanding of the distribution is what ultimately enables us to compute areas and other statistics.
The concept of a probability distribution forms the backbone of the exercise, providing the framework needed to solve for the expected area of the circle and handle further conditions, such as rejecting the furthest shot.