Question

\(X_{1}, X_{2}, \ldots, X_{n}\) are independent identically distributed random variables drawn from a uniform distribution on \([0,1] .\) The random variables \(A\) and \(B\) are defined by $$ A=\min \left(X_{1}, X_{2}, \ldots, X_{n}\right), \quad B=\max \left(X_{1}, X_{2}, \ldots, X_{n}\right) $$ For any fixed \(k\) such that \(0 \leq k \leq \frac{1}{2}\), find the probability \(p_{n}\) that both $$ A \leq k \quad \text { and } \quad B \geq 1-k $$ Check your general formula by considering directly the cases (a) \(k=0,\left(\right.\) b) \(k=\frac{1}{2}\), (c) \(n=1\) and \((\) d) \(n=2\)

Short answer

The probability is given by \( p_n = (1 - (1 - k)^n) * (1 - k^n) \.

Step-by-step solution

Understand the Problem

You need to find the probability that the minimum of several uniformly distributed random variables is less than or equal to a given value k, while the maximum is greater than or equal to 1-k. The variables are defined on the [0, 1] interval, and the conditions are 0 ≤ k ≤ 0.5.

Determine Relevant Probability Conditions

First, note that for the given uniform distribution on [0, 1], the cumulative distribution function (CDF) is F(x) = x for 0 ≤ x ≤ 1. You need to find the conditions such that A = min(X_i) ≤ k and B = max(X_i) ≥ 1-k.

Calculate the Probability for Min Condition

The event A ≤ k implies that at least one of the X_i's is less than or equal to k. The probability that all X_i's are greater than k is (1 - k)^n. So, the probability that A ≤ k is given by P(A ≤ k) = 1 - (1 - k)^n.

Calculate the Probability for Max Condition

The event B ≥ 1 - k implies that at least one of the X_i's is greater than or equal to 1 - k. The probability that all X_i's are less than 1 - k is (1 - (1 - k))^n = k^n. So, the probability that B ≥ 1 - k is given by P(B ≥ 1 - k) = 1 - k^n.

Combine the Conditions using Independence

Since the X_i's are independent, the events A ≤ k and B ≥ 1 - k are also independent. Therefore, the combined probability p_n of both events is the product of their individual probabilities: P(A ≤ k and B ≥ 1 - k) = P(A ≤ k) * P(B ≥ 1 - k) = (1 - (1 - k)^n) * (1 - k^n).

Check the General Formula for Specific Cases

Now solve for specific values of k and n to verify the formula:

Case (a): k = 0

If k = 0, then P(A ≤ 0 and B ≥ 1 - 0) = P(A ≤ 0 and B ≥ 1) = 0.

Case (b): k = 0.5

If k = 0.5, then P(A ≤ 0.5 and B ≥ 0.5) = (1 - (1 - 0.5)^n) * (1 - 0.5^n) = (1 - 0.5^n) * (1 - 0.5^n) = (1 - 0.5^n)^2.

Case (c): n = 1

If n = 1, then P(A ≤ k and B ≥ 1 - k) = (1 - (1 - k)^1) * (1 - k^1). For any 0 ≤ k ≤ 0.5 this becomes k * (1 - k). For k = 0, it's 0. For k = 0.5, it's 0.5 * (1 - 0.5) = 0.25.

Case (d): n = 2

If n = 2, then P(A ≤ k and B ≥ 1 - k) = (1 - (1 - k)^2) * (1 - k^2). For k = 0, it's 0. For k = 0.5, it becomes (1 - 0.25) * (1 - 0.25) = 0.75 * 0.75 = 0.5625.

Key concepts

Uniform Distribution

The uniform distribution is a type of probability distribution where every possible outcome is equally likely within a certain range, known as the support. For example, if we have a uniform distribution on the interval \([0, 1]\), it means every point within this interval is equally likely to be chosen.
A practical example would be rolling a fair die, where each of the six faces has an equal probability of landing face-up. In the context of the exercise, the random variables \(X_{1}, X_{2}, \ldots, X_{n}\) are uniformly distributed on \[0, 1\].
The probability density function (PDF) for a uniform distribution on an interval \[a, b\] is constant, given by \(\frac{1}{b - a}\) for \ a \leq x \leq b\. The CDF, which gives the probability that a uniform random variable takes a value less than or equal to x, is linear within the bounds of the distribution.

Cumulative Distribution Function

The cumulative distribution function (CDF) of a random variable X is a function that describes the probability that X will take on a value less than or equal to x. It's a useful tool for understanding the distribution of values that a random variable can take.
For a uniform distribution on the interval \([0, 1]\), the CDF \(F(x)\) is given by \(F(x) = x\) for \ 0 \leq x \leq 1\. This means the probability that a uniform random variable is less than or equal to x is exactly x. For instance, the probability that a value drawn from a \[0, 1\] uniform distribution is less than 0.5 is 0.5.
In general, the CDF is a non-decreasing function that ranges from 0 to 1 as x moves from \(-\infty\) to \(\infty\). The CDF can be used to calculate probabilities for intervals by finding the difference between the CDF values at the endpoints of the interval.

Independent Random Variables

Independent random variables are those whose occurrences do not affect each other. In other words, the probability of one event occurring does not influence the probability of another event occurring.
Mathematically, two random variables X and Y are independent if the joint probability distribution can be factored into the product of their marginal distributions: \(P(X \in A, Y \in B) = P(X \in A) \cdot P(Y \in B)\), for any sets A and B.
In the given exercise, the variables \(X_{1}, X_{2}, \ldots, X_{n}\) are independent and identically distributed (i.i.d.). This means each variable follows the same uniform distribution and the value of one variable does not depend on the others.

• The independence property is crucial for multiplying probabilities of combined events, like calculating the probabilities of both the minimum being less than or equal to k and the maximum being greater than or equal to \(1 - k\).
• This allows the use of simple product rules in probability calculations, as demonstrated in the step-by-step solution.

Minimum and Maximum of Random Variables

The minimum and maximum of a set of random variables are important concepts. Given several random variables, the minimum is the smallest value among them, and the maximum is the largest.
For uniformly distributed random variables on \[0, 1\]:

• The minimum, \(A = \min(X_{1}, X_{2}, \ldots, X_{n})\), has a CDF of \(F_A(a) = 1 - (1 - a)^n\).
• The maximum, \(B = \max(X_{1}, X_{2}, \ldots, X_{n})\), has a CDF of \(F_B(b) = b^n\).

To solve the exercise, we needed the probability that \(A \leq k\) and \(B \geq 1 - k\). For independent variables, these probabilities can be multiplied:

• \(P(A \leq k) = 1 - (1 - k)^n\)
• \(P(B \geq 1 - k) = 1 - k^n\)

Finally, the combined probability \(P(A \leq k\ and\ B \geq 1 - k)\) is found by multiplying these probabilities, yielding \( (1 - (1 - k)^n) \cdot (1 - k^n) \).