By the argument of weak law of large numbers, if the X_i are i.i.d with mean mu and variance sigma^2, and if M X_n=(X_1+X_2+.....+X_n)/n, then we have an inequality of the form P(|M_ N-mu|>=epsilon)<=(a*sigma^2)/n, for a suiable value of a. 1. If epsilon=0.1, then the value of a is?
To find the value of a, we need to use the Chebyshev's inequality, which is a general form of the weak law of large numbers.
In this case, we are given that M_N = (X_1 + X_2 + ... + X_N)/N, where X_i are i.i.d random variables with mean mu and variance sigma^2.
Chebyshev's inequality states that for any random variable Z with mean mu and variance sigma^2, and any positive constant k, we have:
P(|Z - mu| >= k*sigma) <= 1/k^2
Here, we want to find k such that P(|M_N - mu| >= epsilon) <= (a*sigma^2)/N, where epsilon = 0.1.
Substituting k = epsilon/sigma, we get:
P(|M_N - mu| >= epsilon) <= (sigma^2)/(epsilon^2)
Comparing this with the desired inequality, we can conclude that a = 1/epsilon^2 = 1/(0.1)^2 = 100.
To find the value of 'a', we need to use the weak law of large numbers.
According to the weak law of large numbers, for a sequence of independent and identically distributed (i.i.d) random variables X_i with mean mu and variance sigma^2, the sample mean M_N converges to the population mean mu as the sample size N becomes larger.
The inequality given in the question is:
P(|M_N - mu| >= epsilon) <= (a * sigma^2) / N
To find the value of 'a' when epsilon = 0.1, we need to substitute the given values into the inequality.
P(|M_N - mu| >= 0.1) <= (a * sigma^2) / N
Since the variance is given as sigma^2, we can rewrite it as:
P(|M_N - mu| >= 0.1) <= (a * sigma^2) / N
Now, we can substitute the given value of epsilon (= 0.1) into the inequality:
P(|M_N - mu| >= 0.1) <= (a * sigma^2) / N
Hence, for epsilon = 0.1, the value of 'a' is (a * sigma^2) / N.