91Ó°ÊÓ

A sequence of random variables \(X_{1}, X_{2}, X_{3}, \ldots\) is said to converge stochastically (or in probability) to a random variable X, if for every \(\epsilon > 0\), the probability that the difference of \(X_{n}\) and X exceeds \(\epsilon\) approaches zero, i.e., \(\lim _{n \rightarrow \infty} P\left(\left|X_{n}-X\right|>\epsilon\right)=0\). In this context, we want to check if \(W_{n}\) converges stochastically to \(\mu\).

Chebyshev's Inequality states that, for a random variable with finite expected value \(\mu\) and finite non-zero variance \(\sigma^2\), the probability that the absolute difference of the random variable and its expected value exceeds k times the standard deviation is always less than or equal to \(1/k^2\), for k > 0. Algebraically, \(P\left(\left|X-\mu\right|\geq k \sqrt{\sigma^2}\right) \leq 1/k^2\). Here, we want to show that the absolute difference between \(W_{n}\) and \(\mu\) exceeds \(\epsilon\) with a decreasing probability as \(n\) tends to infinity. Thus, we can implement Chebyshev's Inequality to our exercise by setting \(X = W_{n}\), \(k = \epsilon / \sqrt{b/n^p}\) and \(\sigma^2 = b / n^p\). The inequality becomes: \(P\left(\left|W_{n}-\mu\right|\geq \epsilon\right) \leq (b / n^p) / \epsilon^2\).

As \(P\left(\left|W_{n}-\mu\right|\geq \epsilon\right) \leq (b / n^p) / \epsilon^2\) and since \(b\), \(\epsilon^2\) and \(p\) are constants and \(n\) tends to infinity, the right side of the inequality will approach 0. This means that the probability that the absolute difference of \(W_{n}\) and its expected value \(\mu\) exceeds \(\epsilon\) is also approaching zero. Hence, \(W_{n}\) converges stochastically to \(\mu\).