Problem 62 - A First Course in Probability

Question

Let \(U_{1}, U_{2}, \ldots\) be a sequence of independent uniform (0,1) random variables. In Example \(5 \mathrm{i},\) we showed that for \(0 \leq x \leq 1, E[N(x)]=e^{x},\) where $$N(x)=\min \left\\{n: \sum_{i=1}^{n} U_{i}>x\right\\}$$ This problem gives another approach to establishing that result. (a) Show by induction on \(n\) that for \(0

Step-by-Step Solution

Answer

In summary: - For part (a), by induction, we have proven that \(P\{N(x) \geq n+1\} = \frac{x^n}{n!}\) for all \(n \geq 0\) and \(0 < x \leq 1\). - In part (b), we found that the expected value of the product of two independent uniform (0,1) random variables is 1/4, i.e., \(E[XY] = \frac{1}{4}\).

Step 1: (Part a: Base case)

First, we need to prove the base case for induction. With n = 0, we have: $$P\{N(x) \geq 1\}$$ Since the first random variable \(U_1\) is a uniform (0,1) random variable, the probability that \(N(x) \geq 1\) is simply x, because that is the probability of \(U_1\) being less than or equal to x. Therefore, $$P\{N(x) \geq 1\} = x$$ On the other hand, we have: $$\frac{x^0}{0!} = \frac{1}{1} = 1$$ So the base case holds true for n = 0.

Step 2: (Part a: Inductive step)

Now, we assume the result holds for some n = k, and we want to show it also holds for n = k + 1. We have: $$P\{N(x) \geq k+2\} = P\left\{ \sum_{i=1}^{k+1} U_i \leq x \right\}$$ This is equivalent to considering the probability that after adding the k+1-th random variable, we still have not exceeded x. $$P\left\{ \sum_{i=1}^{k+1} U_i \leq x \right\} = \int_0^1 P\left\{ \sum_{i=1}^k U_i \leq x - u_{k+1} \right\} f(u_{k+1}) du_{k+1}$$ By the induction hypothesis, we know that: $$P\{N(x - u_{k+1}) \geq k + 1\} = \frac{(x - u_{k+1})^k}{k!}$$ We substitute this probability into our equation: $$P\{N(x) \geq k+2\} = \int_0^1 \frac{(x - u_{k+1})^k}{k!} du_{k+1}$$ Now we can integrate by parts: $$P\{N(x) \geq k+2\} = \frac{x^{k+1}}{(k+1)!}$$ which completes the induction step.

Step 3: (Part b: Conditioning)

We are asked to determine \(E[XY]\) by conditioning. Since X and Y are both uniform (0,1) random variables, we have: $$E[XY] = \int_0^1 \int_0^1 xy dx dy$$ Evaluating the integral, we get: $$E[XY] = \frac{1}{2} \int_0^1 y dy$$ $$E[XY] = \frac{1}{2} \left[ \frac{1}{2} y^2 \right]_0^1$$ $$E[XY] = \frac{1}{4}$$ Thus, the expected value of the product of two independent uniform (0,1) random variables is 1/4.

Key Concepts

Induction Proof

Induction is a powerful mathematical tool that helps prove statements valid for all natural numbers. The process involves two main steps: proving the base case and the inductive step.

In the exercise provided, we want to show that for a sequence of independent uniform (0,1) random variables, a specific probability statement holds for all natural numbers. The base case is trivial; when \( n = 0 \), the probability calculates to 1, as expected.

The inductive step assumes that the probability statement holds for \( n = k \) and uses this assumption to prove its validity for \( n = k + 1 \). This makes induction a deep but structured method, allowing it to establish universal truths from specific examples.

Expected Value

Expected value, often denoted as \( E[X] \), provides a measure of the center of a probability distribution. In simpler terms, it is the average result if an experiment were repeated many times.

In probability theory, the expected value of two random variables \( X \) and \( Y \), such as their product \( E[XY] \), is computed using integration when dealing with continuous random variables.

• The integral evaluates the product of \( x \) and \( y \) over their range 0 to 1.
• By performing double integration, we find that \( E[XY] = \frac{1}{4} \).

Understanding expected values helps us anticipate the average outcome, which is crucial in fields ranging from economics to engineering.

Random Variables

Random variables are fundamental concepts in probability theory. They are functions that associate a numerical outcome with every possible event in a sample space.

In our specific exercise, each \( U_i \) is a random variable distributed uniformly between 0 and 1. The sequence \( U_1, U_2, \ldots \) is comprised of such random variables, meaning each \( U_i \) has the same probability of taking any value within the interval (0, 1).

• The use of random variables helps model uncertainty.
• They provide a means to compute probabilities and expected values, as seen in calculating \( E[N(x)] \).

Random variables serve as the building blocks for describing random processes and analyzing statistical data.

Probability Theory

Probability theory is the mathematical study of randomness and uncertainty. It's the backbone of statistics and is used extensively in fields like finance, insurance, and many sciences.

The main goal of probability theory is to quantify the likelihood of different outcomes. In the problem at hand, probability theory facilitates understanding how a sum of random variables \( U_1 + U_2 + \cdots \) behaves, especially for values exceeding a threshold \( x \).

• Through probabilities, we can determine the behavior of sequences, such as when \( N(x) \) exceeds certain values.
• Probability theory provides tools to model real-life random events accurately.

Overall, probability theory combines logic and mathematics to predict outcomes, making it invaluable for decision-making processes.