Chapter 8, Problem 12
(Chebyshev) Assume that X1, X2, . . . , Xn are independent random variables with possibly different distributions and let Sn be their sum. Let
\[ m_k=E(X_k)\\ \sigma^{2}_k=V(X_k)\\ M_k=m_1+m_2+...+m_n \]
Assume that
\[ \sigma^{2}_k<R\\ \forall k \]
Prove that, for any epsilon > 0,
\[ P(|\frac{ S_ n} { n}+\frac{M_n}{n}|<\epsilon )\rightarrow 1 \]
as n approaches infinity
Proof:
The method of proving said entity comes from courses such as real analysis. http://www.milefoot.com/math/calculus/limits/DeltaEpsilonProofs03.htm
How can we use our givens?
We are told that X1, X2, …Xn are independent random variables. We can compute the expected value as follows:
\[ E(S_n)=E(X_1+X_2+X_3+...+X_n)\\ =E(X_1)+E(X_2)_...+E(X_n)\\ =m_1+m_2+...+m_3\\ =M_n \]
Using our givens, we can also compute the variance of S
\[ V(S_n)=V(X_1+X_2+X_3+...+X_n)\\ =V(X_1)+V(X_2)_...+V(X_n)\\ =\sigma^{2}_1+\sigma^{2}_2+...+\sigma^{2}_n\\ =\sum _{ k=1 }^{ n }{ \sigma^{2}_k } \]
Recall the Chebyshev inequality https://www.statlect.com/fundamentals-of-probability/Chebyshev-inequality
\[ P(|X-u|\le k\sigma )\ge 1-\frac { 1 }{ { k }^{ 2 } } \]
We can now proceed to our proof
Apply Chebyshev inequality to our Sn
\[ P(|S_ n+M_n|<k \sigma )\ge \frac { \sigma^{2} }{ { k }^{ 2 } }\\ P(|\frac{ S_ n} { n}+\frac{M_n}{n}|<\frac{k\sigma}{n} )\ge \frac { \sigma^{2} }{ { k }^{ 2 } }\\ P(|\frac{ S_ n} { n}+\frac{M_n}{n}|<\epsilon )\ge 1-\frac { \sigma^{4} }{ { (n\epsilon) }^{ 2 } } \] If you take the limit of the R.H.S as n approaches infinity,you are only left with 1,hence proved. The k plays almost the same role as a delta when doing a delta epsilon proof.