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.

Question

a. If epsilon=0.1, then the value of a is?

b. If we change epsilon = 0.1 to0.1/k, for k>=1, (i.e., if we are interested in k times higher accuracy), how should we change n so that the value of the upper bound does not change from the value calculated in part (a)?

Answer 1

a. To find the value of a, we need to use the weak law of large numbers. The weak law of large numbers states that for a sequence of independent and identically distributed random variables X_i with mean μ and variance σ^2, the sample mean M_n = (X_1 + X_2 + ... + X_n) / n converges in probability to the true mean μ.

Using the weak law of large numbers, we have:

P(|M_n - μ| >= ε) <= (aσ^2) / n

If ε = 0.1, we can substitute it into the inequality:

P(|M_n - μ| >= 0.1) <= (aσ^2) / n

To find the value of a, we need additional information about the distribution of X_i or the value of σ. Without this information, we cannot determine the exact value of a.

b. If we change ε = 0.1 to ε = 0.1/k, we want to find the value of n that keeps the upper bound unchanged.

Using the same inequality as before:

P(|M_n - μ| >= ε) <= (aσ^2) / n

We substitute ε = 0.1/k and solve for n:

P(|M_n - μ| >= 0.1/k) <= (aσ^2) / n

Since we want the upper bound to remain the same, the right side of the inequality should not change. Therefore, we must equate (aσ^2) / n to (aσ^2) / n':

(aσ^2) / n = (aσ^2) / n'

Cancelling σ^2 and a on both sides, we get:

1 / n = 1 / n'

This implies that n = n'.

Therefore, to keep the upper bound unchanged, we do not need to change the value of n when ε is changed to 0.1/k.