I am stuck with this task. Can somebody help me to explain how to solve the problem with Chebyshev's inequality?
The man earns Xn bitcoins on day n, where Xn are independent and identically distributed random variable with mean 6 and Var = 9. Use one-sided and two-sided Chebyshev's inequality to solve estimate the probability that the total amount the man earns in 100 days is less than 510
One sided Chebychev's inequality:
$$P(X\le \mu - a) \le \frac{\sigma^2}{\sigma^2 + a^2}\tag1$$
$$P(X\ge \mu + a) \le \frac{\sigma^2}{\sigma^2 + a^2}\tag2$$
$E(X_i) = 6,E (X) = E(100 X) =100E(X_i) = 600$
$Var (X_i) = 9, Var(X) = Var(nX_i) = nVar(X_i) = 900$
Using (1) $$P(X<510) = P(X\le 600 - 90) \le \frac{900}{900 + 8100} = \frac{1}{10}$$
Using (2)$$P(X>690) = P(X\le 600 + 90) \le \frac{900}{900 + 8100} = \frac{1}{10}$$
Two Sided Chebychev's inequality, you get a relaxed bound
$$P(|X-\mu| > t) \le \frac{\sigma^2}{t^2}\tag3$$
Using (3) $$P(|X-600| > 90) =P(X-600>90) + P(X-600<-90) = P(X>690) + P(X<510) \le \frac{900}{8100} = \frac{1}{9}$$
$$P(X>690) + P(X<510) \le \frac{1}{9}$$
$$P(X<510) \le \frac{1}{9}$$