Question

Let \(X_{i}, i \geqslant 1\), be independent uniform \((0,1)\) random variables, and define \(N\) by $$ N=\min \left\\{n: X_{n}

Short answer

The expected value of \(N\), given by \(f(x)\), is found by conditioning on \(X_1\) and solving the resulting first-order linear differential equation: \[ f(x) = e^{-x^2/2}+1 \]

Step-by-step solution

(Part a) Derive an integral equation for f(x)

Since we know that \(X_0=x\), we can condition the expected value on \(X_1\). That is, we will calculate the expected value based on two cases: 1. If \(X_1 > X_0\) 2. If \(X_1 < X_0\) We have: \[ f(x) = E[N] = \int_0^x E[N | X_1 = u] du \] Notice that: E[N | X_1 = u] = \begin{cases} 1, & \text{if } u < x \\ E[N], & \text{if } u > x \end{cases} Thus: \begin{aligned} f(x) &= \int_0^x E[N | X_1 = u] du \\ & = \int_0^x 1 du + \int_x^1 E[N] du \\ & = x + (1-x)f(x) \end{aligned} This is the integral equation for \(f(x)\).

(Part b) Differentiate both sides of the equation

Now we need to differentiate both sides of this equation with respect to \(x\). We have: \begin{aligned} \frac{d}{dx}f(x) &= \frac{d}{dx}(x + (1-x)f(x)) \\ f'(x) &= 1 + (1-x)f'(x) - f(x) \end{aligned}

(Part c) Solve the resulting differential equation

In order to solve this equation, we will rearrange it: \[ f'(x) - (1-x)f'(x) = 1 - f(x) \] Now, we can express this equation as a first-order linear differential equation: \[ f'(x) = \frac{1-f(x)}{1-x} \] This equation can be solved by multiplying both sides by the integrating factor, which is \(e^{\int (1-x)dx} = e^{x^2/2}\). Then, we have: \[ e^{x^2/2} f'(x) = \frac{e^{x^2/2} (1-f(x))}{1-x} \] Now, integrate both sides with respect to \(x\): \[ f(x) = e^{-x^2/2} \int \frac{e^{x^2/2}(1-f(x))}{1-x} dx + C \] The constant C can be determined by using the boundary condition: \(f(0) = 1\) \[ 1 = e^0 \int \frac{e^{0}(1-f(0))}{1-0} d0 + C \Rightarrow C = 1 \] Thus, the function \(f(x)\) can be expressed as: \[ f(x) = e^{-x^2/2} \int \frac{e^{x^2/2}(1-f(x))}{1-x} dx + 1 \]

(Part d) Argue the given probability function

We are given that: \[ P\{N \geqslant k\}=\frac{(1-x)^{k-1}}{(k-1) !} \] This probability represents the probability that there have been at least \(k-1\) values in the sequence that are greater than their preceding values. Since we have i.i.d. uniform random variables in the sequence, the probability that \(k-1\) are greater than their predecessors is a direct consequence of the memoryless property.

(Part e) Obtain f(x) using part (d)

To find \(f(x)\), we know that the expected value is given by: \[ f(x) = E[N] = \sum_{k=1}^{\infty}kP\{N=k\} \] Since we have \(P\{N \geqslant k\}\) and we know that probabilities sum to 1, we can find \(P\{N=k\}\): \[ P\{N=k\} = P\{N \geqslant k\} - P\{N \geqslant k+1\} = \frac{(1-x)^{k-1}}{(k-1) !} - \frac{(1-x)^{k}}{k !} \] Then, we compute f(x): \begin{aligned} f(x) &= \sum_{k=1}^{\infty}k \left( \frac{(1-x)^{k-1}}{(k-1) !} - \frac{(1-x)^{k}}{k !} \right) \\ & = \sum_{k=1}^{\infty} \left( k \frac{(1-x)^{k-1}}{(k-1) !} - (k-1) \frac{(1-x)^{k-1}}{(k-1) !} \right) \\ & = \sum_{k=1}^{\infty} \frac{(1-x)^{k-1}}{(k-1) !} \left( k - (k-1) \right) \\ & = \sum_{k=1}^{\infty} \frac{(1-x)^{k-1}}{(k-1) !} \\ & = e^{-x^2/2}+1 \end{aligned} Thus, the expected value of \(N\), given by \(f(x)\), is: \[ f(x) = e^{-x^2/2}+1 \]

Key concepts

Understanding Expected Value

When delving into probability models, understanding the concept of expected value is crucial. Expected value, often denoted as E[X], is a cornerstone in probability and statistics, representing the average outcome we would expect if an experiment were repeated an infinite number of times.

Think of it as a weighted average of all possible values that a random variable can take, with 'weights' being the probabilities of each outcome. For example, if we flip a fair coin, the expected value of the number of heads in one flip is 0.5, since there's a 50% chance for heads and 50% for tails, and we average those probabilities over the two possible outcomes, head (1) and tail (0).

In the context of the exercise, the expected value E[N] represents the average position N in a sequence where a uniform random variable is less than its predecessor for the first time. This application illustrates how the concept is broader than just simple games of chance and can be applied to sequencing and ordering problems as well.

The Role of Integral Equations in Probability

An integral equation is a relationship between an unknown function and its integration. Integral equations often appear in the field of differential equations, but they have a special place in probability models, too. They are significant for problems where we want to describe a system whose state changes over time or space and the changes are cumulative in nature.

In the provided exercise, the integral equation is derived by conditioning on an event, showcasing how probability models can help us set up equations that define expected values. Here the aim is to find a function that describes the expected value of N, where the integral sign indicates summing up the contributions from all possible values that the next random variable can take. The integration bounds being from zero to x results from the uniform nature of the random variables, which only take values in the range (0,1).

Such equations are powerful as they amalgamate continuous change (through integration) and the probabilistic behavior of random variables, allowing us to understand the potential outcomes of complex processes.

Uniform Random Variables Unveiled

The concept of uniform random variables is particularly interesting in probability theory. A uniform random variable is a type of continuous random variable with all intervals of the same length within its range being equally probable. Essentially, it's the 'fair' random variable for continuous outcomes, just as a fair die is for discrete results.

In our exercise, the uniform random variables are defined between 0 and 1, which means any number in this interval is just as likely to occur as any other. This uniformity simplifies many calculations in probabilities, as seen in part (d) of the exercise where the memoryless property becomes apparent. The uniform distribution's simplicity can sometimes mislead, but it illustrates fundamental aspects of probability and is an excellent introduction to more complex distributions.

It's important to note that when dealing with uniform random variables, their straightforward nature often lends to more transparent interpretations of probability models and thus are a fundamental concept for students to grasp in the study of probability.