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An individual traveling on the real line is trying to reach the origin. However, the larger the desired step, the greater is the variance in the result of that step. Specifically, whenever the person is at location \(x\), he next moves to a location having mean 0 and variance \(\beta x^{2}\). Let \(X_{n}\) denote the position of the individual after having taken \(n\) steps. Supposing that \(X_{0}=x_{0}\), find (a) \(E\left[X_{n}\right]\); (b) \(\operatorname{Var}\left(X_{n}\right)\)

Short Answer

Expert verified
In summary, we find that the expected position of the individual after \(n\) steps is \(E[X_n] = x_0\) for \(n \geq 1\). The variance after \(n\) steps is given by the recursive formula \(\operatorname{Var}(X_n) = \sum_{i=0}^{n-1} \beta (X_i)^2\) for \(n \geq 1\).

Step by step solution

01

Determine the given information

For an individual starting at position \(x_0\) and taking \(n\) steps on the real line, we are given that while being at position \(x\), he moves to a location with mean 0 and variance \(\beta x^{2}\). Let \(X_n\) denote the position of the individual after n steps.
02

(a) Find \(E[X_n]\)

To find the expected position, we need to compute the expected value of the function \(X_{n}\). Using the linearity of expected values, we can find the expectation for each step and then sum them up over \(n\) steps. So for the first step: \[E[X_1] = E[x_0 + M_1] = E[x_0] + E[M_1] = x_0,\] where \(M_1\) is the movement in the first step with mean 0. For the subsequent steps, \[E[X_{n}] = E[X_{n-1} + M_n],\] where \(M_n\) represents the movement in the \(n\)th step. Since \(E[M_n] = 0\), we have: \[E[X_{n}] = E[X_{n-1}],\] Which results in a constant expected position throughout the process: \[E[X_n] = x_0 \quad \text{for} \quad n \geq 1.\]
03

(b) Find \(\operatorname{Var}(X_n)\)

To find the variance, we can use the property of variances that states \(\operatorname{Var}(X+Y) = \operatorname{Var}(X) + \operatorname{Var}(Y)\), when \(X\) and \(Y\) are independent. In this case, \(X_n = X_{n-1} + M_n\). Let's compute the variance for the first step: \[\operatorname{Var}(X_1) = \operatorname{Var}(x_0 + M_1) = \operatorname{Var}(M_1) = \beta x_0^{2}.\] Now, let's compute the variance for the next steps: \[\operatorname{Var}(X_n) = \operatorname{Var}(X_{n-1} + M_n) = \operatorname{Var}(X_{n-1}) + \operatorname{Var}(M_n) = \operatorname{Var}(X_{n-1}) + \beta(X_{n-1})^2.\] To solve this recursively, we already know that \(\operatorname{Var}(X_1) = \beta x_0^{2}\). Therefore, we can calculate the subsequent variances as follows: \[\operatorname{Var}(X_n) = \sum_{i=0}^{n-1} \beta (X_i)^2 \quad \text{for} \quad n \geq 1.\] To get a sense of the recursive formula, you can compute the first few variances explicitly as: \[\operatorname{Var}(X_2) = \operatorname{Var}(X_1) + \beta (X_1)^2 = \beta x_0^2 + \beta(E[X_1])^2\] \[\operatorname{Var}(X_3) = \operatorname{Var}(X_2) + \beta (X_2)^2 = \beta x_0^2 + \beta (E[X_1])^2 + \beta (E[X_2])^2\] And so on. Overall, we derived both the expected position \(E[X_n]\) and variance \(\operatorname{Var}(X_n)\) after \(n\) steps using the provided information about the relationship between step size and variance.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean and Variance
Understanding the concept of mean and variance is essential when dealing with probability models. In simple terms, the mean represents the average or expected value of a set of data. For a random variable, it is the sum of every possible value, each multiplied by its probability. In this random walk problem, since the movement has a mean of 0, the mean of the position after any number of steps remains constant and equals the starting position, denoted as \(x_0\).

Variance measures how much the values of a random variable differ from its mean. In mathematical terms, it is the expected value of the squared deviations from the mean. In our random walk, as the individual takes steps, each with variance \(\beta x^2\), it showcases how the more steps taken, the more "spread out" the probable positions become.

It's crucial in probability models to determine both the mean and variance, especially when assessing how different outcomes might be distributed around an expected value. Knowing these can help predict potential outcomes more accurately.
Random Walk Problem
A random walk is a mathematical model used to describe a path consisting of a succession of random steps. In this exercise, it's about an individual on the real line attempting to reach the origin, but the path taken is anything but straightforward.

  • The individual can travel from one position to another, but the variance or uncertainty of each step depends on where they currently are.
  • The mean position after each step is zero, so, on average, the individual doesn't move closer or further from the origin. However, the variance, calculated as \(\beta x^2\), increases as they move, which means more scattering around the origin.
This kind of problem is vital in fields like finance for modeling stock prices or in physics to describe molecules' movements in space. Understanding a random walk helps in predicting eventual position distributions after a significant number of steps.
Step-by-Step Solution
Getting through a random walk problem effectively requires breaking it down systematically. Let's look deeper into the step-by-step approach:

First, identify all given data. Here, you know the initial position \(X_0 = x_0\), and each step has a mean of 0 and variance \(\beta x^2\). Next, for part (a), compute \(E[X_n]\), the expected position. This is streamlined by the fact that each step has a mean of 0, keeping the initial mean constant all through:

\[E[X_n] = x_0, \quad n \geq 1.\]

For part (b), determine \(\operatorname{Var}(X_n)\). Here, apply the variance property: variances add up when independent. From the first step:

\[\operatorname{Var}(X_1) = \beta x_0^2.\]

Then use recursion for subsequent steps:
\[\operatorname{Var}(X_n) = \operatorname{Var}(X_{n-1}) + \beta(X_{n-1})^2.\]

This step-by-step process empowers you to solve complex problems by treating each step with equal importance and building understanding gradually, which is invaluable in problem-solving situations in math and science.

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Most popular questions from this chapter

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

Independent trials, resulting in one of the outcomes \(1,2,3\) with respective probabilities \(p_{1}, p_{2}, p_{3}, \sum_{i=1}^{3} p_{i}=1\), are performed. (a) Let \(N\) denote the number of trials needed until the initial outcome has occurred exactly 3 times. For instance, if the trial results are \(3,2,1,2,3,2,3\) then \(N=7\) Find \(E[N]\). (b) Find the expected number of trials needed until both outcome 1 and outcome 2 have occurred.

Prove that if \(X\) and \(Y\) are jointly continuous, then $$ E[X]=\int_{-\infty}^{\infty} E[X \mid Y=y] f_{Y}(y) d y $$

In the list example of Section \(3.6 .1\) suppose that the initial ordering at time \(t=0\) is determined completely at random; that is, initially all \(n !\) permutations are equally likely. Following the front-of-the-line rule, compute the expected position of the element requested at time \(t\). Hint: To compute \(P\left\\{e_{j}\right.\) precedes \(e_{i}\) at time \(\left.t\right\\}\) condition on whether or not either \(e_{i}\) or \(e_{j}\) has ever been requested prior to \(t\).

The number of coins that Josh spots when walking to work is a Poisson random variable with mean 6 . Each coin is equally likely to be a penny, a nickel, a dime, or a quarter. Josh ignores the pennies but picks up the other coins. (a) Find the expected amount of money that Josh picks up on his way to work. (b) Find the variance of the amount of money that Josh picks up on his way to work. (c) Find the probability that Josh picks up exactly 25 cents on his way to work.

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