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.
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)?
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.
To find the value of "a" when epsilon=0.1, we need to use the weak law of large numbers inequality. The weak law states that for a sequence of independent and identically distributed random variables (X_i), if M_X_n = (X_1 + X_2 + ... + X_n) / n, then:
P(|M_n - mu| >= epsilon) <= (a * sigma^2) / n
where "mu" is the mean of X_i and "sigma^2" is the variance of X_i.
a. For epsilon=0.1, we need to find the value of "a" that satisfies the inequality. Since the exact value of "a" is not given in the question, we'll solve for it using the given information.
In this case, we can rewrite the inequality as:
P(|M_n - mu| >= 0.1) <= (a * sigma^2) / n
Since "mu" and "sigma^2" are given, we need to calculate the right-hand side of the inequality.
b. Now, let's consider changing epsilon to 0.1/k, where k is a positive integer greater than or equal to 1. We want to find how n should change so that the value of the upper bound remains the same as calculated in part (a).
When changing epsilon to 0.1/k, the inequality becomes:
P(|M_n - mu| >= 0.1/k) <= (a * sigma^2) / n
To keep the value of the upper bound unchanged, we need to choose a value of n that compensates for the change in epsilon.
In this case, we need to solve for the new value of n that satisfies the equation:
(a * sigma^2) / n = (a * sigma^2) / (n/k)
Simplifying the equation:
n = n/k
n = k
Therefore, to maintain the same upper bound value, we should set n = k when changing epsilon to 0.1/k.