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The expected value of a random variable \(X\) is defined as \(E[X] = \sum_i x_i P(x_i)\), where \(x_i\) are the possible values of the random variable and \(P(x_i)\) are the probabilities associated with these values. The variance of a random variable \(X\) is defined as \(\operatorname{Var}(X) = E[X^2] - (E[X])^2\).

Given the information in the problem, we can find the expected location at the nth step \(E[X_n]\) using conditional expectation. We know that the mean of the next step, \(E[X_{n+1} | X_n]\), is 0. Therefore, \(E[X_{n+1} | X_n] = 0\) Now we can write the expression for \(E[X_n]\) using conditional expectation: \(E[X_n] = E[E[X_n | X_{n-1}]]\) \(E[X_n] = E[0]\) \(E[X_n] = 0\) So, the expected location after n steps, \(E[X_n]\), is 0.

Now we need to find the variance of the nth step, \(\operatorname{Var}(X_n)\). Using the definition of variance, we know that: \(\operatorname{Var}(X_n) = E[X_n^2] - (E[X_n])^2\) Since we know that \(E[X_n] = 0\), we just need to find \(E[X_n^2]\). Using conditional expectation, we can write: \(E[X_n^2] = E[E[X_n^2 | X_{n-1}]]\) Given that the variance of the individual's position at location \(x\) is \(\beta x^2\), we have: \(E[X_n^2] = E[\operatorname{Var}(X_n | X_{n-1}) + (E[X_n | X_{n-1}])^2]\) Since \(E[X_n | X_{n-1}] = 0\), we have: \(E[X_n^2] = E[\operatorname{Var}(X_n | X_{n-1})]\) \(E[X_n^2] = E[\beta X_{n-1}^2]\) Now, by using the law of iterated expectations, \(E[X_n^2] = \beta E[X_{n-1}^2] = \beta^n E[X_0^2]\) Now we can find the variance of the nth step: \(\operatorname{Var}(X_n) = E[X_n^2] - (E[X_n])^2\) \(\operatorname{Var}(X_n) = \beta^n E[X_0^2] - 0\) \(\operatorname{Var}(X_n) = \beta^n E[X_0^2]\) Thus, the variance of the individual's position after n steps, \(\operatorname{Var}(X_n)\), is \(\beta^n E[X_0^2]\). To summarize: (a) The expected location after n steps, \(E[X_n]\), is 0. (b) The variance of the individual's position after n steps, \(\operatorname{Var}(X_n)\), is \(\beta^n E[X_0^2]\).