1. a) E[H] = E[1/n ∑(i=1 to n) Xi] = (1/n) ∑(i=1 to n) E[Xi] = (1/n) * n * E[Xi] = E[Xi] = h
b) Var(H) = Var(1/n ∑(i=1 to n) Xi) = (1/n^2) ∑(i=1 to n) Var(Xi) = (1/n^2) * n * Var(Xi) = Var(Xi) / n
Since Var(Xi) ≤ 4 (given information),
Var(H) ≤ 4 / n
2. We want the standard deviation of H to be at most 0.01. The standard deviation is the square root of the variance, so:
0.01 ≥ √(Var(H))
Taking square on both sides:
0.0001 ≥ Var(H)
From part (1b), we know that Var(H) ≤ 4 / n. Substituting this value, we have:
0.0001 ≥ 4 / n
Solving for n:
n ≥ 4 / 0.0001 = 40,000
Therefore, the smallest possible value of n is 40,000.
3. The Chebyshev inequality states that for any constant k > 0, the probability that a random variable X deviates from its mean by more than k standard deviations is at most 1/k^2. In this case, we want to be at least 96% sure that our estimate is within 0.02 of the true mean h.
Let k be the number of standard deviations. We want 1/k^2 ≥ 0.04 (96% = 0.96). Solving for k:
k^2 ≤ 1 / 0.04 = 25
Taking the square root on both sides:
k ≤ 5
Since the standard deviation is the square root of the variance, we have:
k ≤ σ(H) / √n
Substituting the given information Var(H) ≤ 4 / n, and σ(H) = √Var(H):
5 ≤ √(4/n) / √n
Squaring both sides:
25 ≤ 4/n
Solving for n:
n ≤ 4/25 = 0.16
Rounding up to the nearest whole number, the minimum value of n is 1.
4. If X is uniformly distributed on [h - 3, h + 3], for some unknown h, the sample mean H will asymptotically follow a normal distribution with mean h and variance 36/n (according to the Central Limit Theorem).
A 95% confidence interval for h is given by:
H ± (1.96 * √(Var(H)))
Since Var(H) = 36/n, the confidence interval becomes:
H ± (1.96 * √(36/n))
Substituting H = h, we obtain:
h ± (1.96 * √(36/n))
This is the most appropriate expression for a 95% confidence interval for h.