Question

Let \(X_{i}, i \geqslant 0\) be independent and identically distributed random variables with probability mass function $$ p(j)=P\left[X_{i}=i\right\\}, \quad j=1, \ldots, m, \quad \sum_{j=1}^{m} P(j)=1 $$ Find \(E[N]\), where \(N=\min \left[n>0: X_{n}=X_{0}\right\\}\)

Short answer

The short answer for the expected value of \(N\) in the given situation is: \[ E[N] = \frac{1}{P(j)} \] where \(P(j)\) is the probability of picking outcome \(j\) as the value of the random variable when \(X_0 = j\).

Step-by-step solution

Determine the Probability Mass Function for \(N\)

If \(X_0 = j\), then to find \(P(N=n)\), we must calculate the probability of not choosing \(j\) in the first \(n-1\) observations and then choosing \(j\) on the \(n\)th observation. Due to the independent and identically distributed nature of the observations, this can be solved as follows: \[ P(N=n) = \,(1 - P(j))^{n-1} \times P(j),\quad n=1,2,3, \cdots \]

Determine \(E[N]\) using the Definition of Expectation

Now that we have the Probability Mass Function for \(N\), we can calculate the expectation \(E[N]\). It's given by: \[ E[N] = \sum_{n=1}^{\infty} nP(N=n) = \sum_{n=1}^{\infty} n \,[\,(1 - P(j))^{n-1} \times P(j)\,] \] This equation is a sum of an infinite sequence which is known to be equivalent to: \[ E[N] = \frac{1}{P(j)} \] Hence, the expected value of \(N\) is the reciprocal of the probability of picking \(j\) as the outcome of the random variable when \(X_0 = j\).