9.3.2
Let \({X_k }\), \(1 ≤ k ≤ n\), be a sequence of independent random variables, all with mean \(\mu = 0\) and variance \(\sigma^2 = 1\) , and let \(S_n, S_n^*\) and \(A_n\) be their sum, standardized sum, and average, respectively. Verify directly that \(_n^∗ = \frac{S_n}{\sqrt{n}} = \sqrt{n}A_n\).
Example 9.7 pg 366 shows us the \(S_n^* = \frac{S_n-n\mu}{\sqrt{n}\sigma}\), so: \[S_n^* = \frac{S_n -n\mu}{\sqrt{n}} = \frac{S_n -n*0}{\sqrt{n}} = \frac{S_n}{\sqrt{n}}\]
The Average, \(A_n\) is given as the sum/n, or \(\frac{S_n}{n}\)
Solve the previous result for \(S_n = \sqrt{n}S_n^*\) and substitue in: \[A_n = \frac{\sqrt{n}S_n^*}{n} = \frac{\sqrt{n}S_n^*}{\sqrt{n}*\sqrt{n}} = \frac{S_n^*}{\sqrt{n}}\] Solving the result \(A_n = \frac{S_n^*}{\sqrt{n}}\), it follows that …
\[S_n^* = \sqrt{n}A_n\]