By the argument in the last video, if the Xi are i.i.d. with mean μ and variance σ2 , and if Mn=(X1+⋯+Xn)/n , then we have an inequality of the form
P(|Mn−μ|≥ϵ)≤aσ2n,
for a suitable value of a .
a) If ϵ=0.1 , then the value of a is:
unanswered
b) If we change ϵ=0.1 to ϵ=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 (a)?
n should
stay the same
increase by a factor of k
increase by a factor of k2
decrease by a factor of k
none of the above
a). a = 100
b). increase by a factor of k^2
Well, changing the accuracy from ϵ = 0.1 to ϵ = 0.1/k means we want k times higher accuracy. In order for the value of the upper bound to remain the same, we need to decrease the probability, which means making it smaller.
Since n appears in the denominator of the inequality, we can decrease n to make the inequality smaller. Therefore, the correct answer is to decrease n by a factor of k.
So, the answer is: decrease by a factor of k. But keep in mind, changing n alone might not be enough to ensure the upper bound does not change, as other factors can also affect it.
a) If ϵ=0.1, then the value of a can be determined by finding the appropriate probability level for a given confidence interval. Let's calculate it step-by-step:
1. Start with the equation we have:
P(|Mn - μ| ≥ ϵ) ≤ aσ²/n
2. We want to find the value of a, so rearrange the equation:
a = P(|Mn - μ| ≥ ϵ) * n/σ²
3. We need to find the probability P(|Mn - μ| ≥ ϵ). This probability can be approximated using the Central Limit Theorem. For large enough n, the random variable (Mn - μ)/σ follows a standard normal distribution.
4. The probability can be expressed as:
P(|Mn - μ| ≥ ϵ) ≈ 2(1 - Φ(ϵ/σ))
Here, Φ(z) represents the cumulative distribution function (CDF) of a standard normal distribution.
5. Calculate the value of Φ(ϵ/σ). You can use a standard normal distribution table or a statistical software to find this value.
6. Substituting the value of Φ(ϵ/σ) into the equation, we find:
a ≈ 2(1 - Φ(ϵ/σ)) * n/σ²
Now, you can substitute the values of ϵ=0.1, σ², and n to compute the value of a.
b) If we change ϵ=0.1 to ϵ=0.1/k, we are interested in k times higher accuracy. To find the new value of n such that the upper bound remains the same, we can use the equation:
P(|Mn - μ| ≥ ϵ_new) ≤ aσ²/n_new
We want to keep the upper bound the same, which means the probability term should remain the same. So:
P(|Mn - μ| ≥ ϵ) ≈ P(|Mn - μ| ≥ ϵ_new)
2(1 - Φ(ϵ/σ)) = 2(1 - Φ(ϵ_new/σ))
Since ϵ_new = ϵ/k, we have:
2(1 - Φ(ϵ/σ)) = 2(1 - Φ((ϵ/k)/σ))
Set these two expressions equal to each other and solve for n_new. Keeping n the same is not the correct answer. The correct answer is to increase n by a factor of k.
To find the value of 'a' in the inequality given by the expression P(|Mn-μ|≥ϵ)≤aσ²/n, we need to refer to the argument presented in the last video. Unfortunately, as an AI text-based bot, I do not have access to video content or its specific arguments. However, I can provide you with a general explanation of how to approach this problem.
In order to determine the value of 'a' when ϵ = 0.1, we need to utilize the properties of the random variables Xi being independently and identically distributed (i.i.d.) with a mean μ and variance σ². The quantity Mn represents the average of the first 'n' observations (X1 + X2 + ... + Xn)/n.
The inequality P(|Mn-μ|≥ϵ)≤aσ²/n states that the probability of the absolute difference between Mn and μ exceeding ϵ is less than or equal to a constant value multiplied by σ² divided by n.
To find the value of 'a', we typically rely on results from probability theory or related statistical literature. These results, such as inequalities involving the cumulative distribution function for the standardized variable Z, can provide bounds on the probability distribution of Mn around its mean μ.
Since the exact argument from the video is not provided, it is challenging to determine the specific value of 'a'. Generally, 'a' will depend on properties of the underlying random variables Xi and the specific distribution they follow. You may need to refer to the specific video or consult the relevant literature to find the answer.
Moving on to part b) of the question, where ϵ is changed to ϵ = 0.1/k for k ≥ 1, and we are interested in k times higher accuracy. The question asks how we should change 'n' so that the value of the upper bound does not change from part a).
To address this, we need to consider the effect of changing ϵ to ϵ = 0.1/k. This change in ϵ implies a smaller allowable difference between Mn and μ, indicating a higher level of accuracy or precision. As ϵ gets smaller, we usually need a larger sample size 'n' to maintain the same level of confidence.
In this case, the value of 'n' should increase by a factor of k^2. This increase compensates for the higher accuracy required, ensuring that the value of the upper bound remains the same as in part a).
Therefore, the correct answer to part b) is "increase by a factor of k^2".