Question

Consider a random variable X such that P(x=1)=1/4 and P(x=0)=3/4. E(x)=1/4 and standard deviation is sqrt(3)/4. If X1, X2,..., Xn is a random sample of the population X, then lim n -> infinity of P(X1+X2+...+Xn/n > C) = 0 for every C>1/4.

Answer 1

Answer on Question #42453 - Math - Statistics and Probability

Consider a random variable $X$ such that $P(x=1)=1/4$ and $P(x=0)=3/4$. $E(x)=1/4$ and standard deviation is $\operatorname{sqrt}(3)/4$. If $X1, X2, \ldots, Xn$ is a random sample of the population $X$, then $\lim n \to \infty$ and $\lim n > C$ for every $C > 1/4$.

Proof

Let $\bar{X}_n = \frac{X_1 + X_2 + \cdots + X_n}{n}$.

Compute $E(\bar{X}_n) = \frac{E(X_1) + E(X_2) + \cdots + E(X_n)}{n} = \frac{nE(X_1)}{n} = E(X_1) = \frac{1}{4}$,

$$Var(\bar{X}_n) = Var\left(\frac{X_1 + X_2 + \cdots + X_n}{n}\right) = \frac{1}{n^2} Var(X_1 + X_2 + \cdots + X_n) = \frac{n\sigma^2}{n^2} = \frac{\sigma^2}{n} = \frac{3}{16n},$$

Using Chebyshev's inequality on $\bar{X}_n$ results in $P\{|\bar{X}_n - E(\bar{X}_n)| \geq \varepsilon\} \leq \frac{\sigma^2}{n\varepsilon^2}$.

$$P\{|\bar{X}_n - E(\bar{X}_n)| \geq \varepsilon\} = P\left\{(\bar{X}_n \geq \varepsilon + E(\bar{X}_n)) \cup (\bar{X}_n \leq E(\bar{X}_n) - \varepsilon)\right\} \leq \frac{\sigma^2}{n\varepsilon^2}.$$

$0 \leq |\text{non-negativity of probability}| \leq P\{\bar{X}_n \geq \varepsilon + E(\bar{X}_n)\} \leq |\text{monotonicity of probability}| \leq$

$$\leq P\left\{(\bar{X}_n \geq \varepsilon + E(\bar{X}_n)) \cup (\bar{X}_n \leq E(\bar{X}_n) - \varepsilon)\right\} \leq |\text{Chebyshev's inequality on } \bar{X}_n| \leq \frac{\sigma^2}{n\varepsilon^2}$$

We obtain $P\{\bar{X}_n > C\} \leq \frac{\sigma^2}{n\varepsilon^2}$, for every $C > E(\bar{X}_n) = \frac{1}{4}$.

Pass to the limit and consider $0 \leq \lim_{n \to \infty} P\left(\frac{X_1 + X_2 + \cdots + X_n}{n} > C\right) \leq \lim_{n \to \infty} \frac{\sigma^2}{n\varepsilon^2} = 0, \varepsilon, C, \sigma^2$ are fixed.

By squeeze theorem we conclude

$$\lim_{n \to \infty} P\left(\frac{X_1 + X_2 + \cdots + X_n}{n} > C\right) = 0 \text{ for every } C > \frac{1}{4}.$$

Q.E.D

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