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was revising my stuffs for my stochastics exams and came across this question that I couldn't figure my way around..


Let $X_1,\ldots,X_n$ be independent, identically distributed random variables with $E(X_1) =a$ and

$$S_n = \frac{1}{n} \sum_{i = 1}^n X_i$$

Using the Chebychev inequality, give the smallest possible value of $x$, where $\mathbb{P}(\left | S_{100} - a \right | \geq x) \leq 0.01$, in the case where $X_{1}$ with the parameters $p \in [0,1]$ is:

a) binomially distributed ($B(10,p)$)

b) geometrically distributed


I gathered that $0.01 =\frac{Var(X))}{x^2}$ owing to the Chebychev inequality and figured that if I could derive $Var(X)$, I could then get $x$. And using the given distributions (binomial or geometric), I attempted to solve for $Var(X) = E(X^2) - (E(X))^2$, although I couldn't proceed any further with the finding of $E(X^2)$.

Although first and foremost, am I heading off into the right track? If so, how should I go about finding $E(X^2)$? Thanks for the help, as always!

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you need to calculate $E(X^2)$ based on underlying distribution of X. For example for binomial distribution check this link proofwiki.org/wiki/Variance_of_Binomial_Distribution –  karakfa Jul 17 '12 at 15:46

1 Answer 1

up vote 1 down vote accepted

I think this is the right track. Since $S_{n}$ is a sum of iid r.v.s we have

\begin{equation} \text{E}(S_n) = \frac{1}{n} \text{E}(X_1+\ldots+X_n) =\frac{1}{n}\left[ \text{E}(X_1)+\ldots + \text{E}(X_n) \right] = \text{E}(X_1) = a \end{equation} This can be used to relate $a$ to the mean of the distributions you want. Similarly,

\begin{equation} \text{Var}(S_n) = \text{Var}\left(\frac{X_1+\ldots+X_n}{n}\right) = \frac{1}{n^2} \text{Var}(X_1+\ldots+X_n) = \frac{n}{n^2} \text{Var}(X_1) = \frac{\text{Var}(X_1)}{n} \end{equation} This follows since, being iid, the $X_i$ are uncorrelated. Now, noting that $0.01 = 1/n$ the result becomes \begin{equation} x^2 = \text{Var}(X_1) \end{equation} Thus,

1) For $X_i \sim Bin(10,p)$, $\text{Var}(X_1) = 10 p(1-p)$

2) For $X_i \sim Geom(p)$, $\text{Var}(X_1) = (1-p)/p^2$

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