By the argument of the weak law of large numbers, if the X_i are i.i.d with mean mu and variance sigma^2, and if M_n=(X_1+....+X_n)/n, then we have an inequality of the form P(|M_n-mu|>=epsilon)<=(a*sigma^2)/n, for a suitable value of a.
1. If epsilon=0.1, then the value of a is?
2. If we change epsilon=0.1 to epsilon=0.1/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 1. n should?
a. stay the same
b. increase by a factor of k
c. increase by a factor of k^2
d. decrease by a factor of k
e. none of the above
a=100
If a = 100, then for any given epsilon:
P(|M_n - mu| >= epsilon) <= (100 * sigma^2) / n.
This is the inequality derived from the weak law of large numbers, with a value of a equal to 100.
1. To determine the value of a when ε = 0.1, we need to use the weak law of large numbers inequality. The inequality states that P(|M_n - μ| ≥ ε) ≤ (aσ^2) / n.
Since ε = 0.1, we have P(|M_n - μ| ≥ 0.1) ≤ (aσ^2) / n.
The exact value of a depends on the specific distribution and properties of X_i. However, if we assume X_i follows a normal distribution, we can use the standard z-score table to find an appropriate value for a.
For ε = 0.1, a commonly used value is a = 2.576. This corresponds to a 99% confidence level. Using this value, the inequality becomes:
P(|M_n - μ| ≥ 0.1) ≤ (2.576σ^2) / n.
2. If we change ε from 0.1 to 0.1/k, we need to determine how n should change in order to keep the upper bound the same.
The weak law of large numbers inequality states that P(|M_n - μ| ≥ ε) ≤ (aσ^2) / n.
Substituting ε = 0.1/k into the inequality, we get:
P(|M_n - μ| ≥ 0.1/k) ≤ (aσ^2) / n.
To keep the upper bound the same, we want the right-hand side of the inequality to be unchanged. This means (aσ^2) / n should remain the same.
Since the only variable that changes is ε, we can conclude that n should remain the same (option a).
Therefore, the answer is: a. stay the same.
For part 1, if we take epsilon=0.1, then
P(|M_n-mu|>=0.1) <= (a*sigma^2)/n.
Since the weak law of large numbers states that M_n converges to mu in probability as n approaches infinity, we can assume that as n becomes larger, the probability of |M_n - mu| being greater than or equal to epsilon becomes smaller. Therefore, we can choose a to be a small positive constant such as 1.
So, for part 1, the value of a is 1.
For part 2, if we change epsilon to epsilon/k, we want to find the value of n such that
P(|M_n-mu|>=epsilon/k) <= (a*sigma^2)/n.
Since we want the upper bound to remain the same, we need to adjust n accordingly. This means that
P(|M_n-mu|>=0.1) <= (a*sigma^2)/n,
and
P(|M_n-mu|>=0.1/k) <= (a*sigma^2)/n.
To make the upper bound remain the same, the probability of |M_n - mu| being greater than or equal to 0.1/k should be the same as the probability of |M_n - mu| being greater than or equal to 0.1. Since the probabilities are decreasing as n becomes larger, we can conclude that the value of n should stay the same.
Therefore, for part 2, the value of n should stay the same. The answer is (a) stay the same.