Question

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

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

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.

(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: \[E[X_n] = x_0 \quad \text{for} \quad n \geq 1.\]

(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.