Let
Yn=∑i=1n(−1)iXi=−X1+X2−X3+⋯,
where the Xi are i.i.d., with E[Xi]=0, and var(Xi)=4.
Is
1nYn
approximately normal? Choose the most appropriate response:
No.
Yes, because the {Yi}i=1∞ are i.i.d., so we can directly apply the Central Limit Theorem.
Yes, because the {(−1)iXi} are i.i.d., so we can directly apply the Central Limit Theorem.
Yes, because Yn/n is the sum of two independent approximately normal random variables.
Yes, because E[Yi]=0 and var(Yi) is finite, so we can directly apply the Weak Law of Large Numbers.
(i)
(ii)
(iii)
(iv)
(v)
unanswered
Find the variance of Yn.
var(Yn)= unanswered
Find P(Y100≥20), approximately.
P(Y100≥20)≈ unanswered
To determine if 1/nYn is approximately normal, we need to analyze the characteristics of Yn.
First, let's find the mean and variance of Yn:
The mean of Yn can be calculated as E[Yn] = E[∑i=1n(−1)iXi].
Since the Xi are i.i.d. with mean E[Xi] = 0, the terms −Xi and Xi will cancel each other out on average. Therefore, E[Yn] = 0.
To find the variance of Yn, we can use the properties of the variance:
var(Yn) = var(∑i=1n(−1)iXi).
Given that the Xi are i.i.d. with variance var(Xi) = 4, we can use the fact that var(aX) = a^2 * var(X):
var(Yn) = var(−X1 + X2 − X3 + ⋯ + (−1)nXn)
= var(−X1) + var(X2) + var(−X3) + ⋯ + var(−Xn)
= 1 * var(X1) + 1 * var(X2) + 1 * var(X3) + ⋯ + 1 * var(Xn)
= n * var(X1)
= n * 4.
Therefore, var(Yn) = 4n.
Now, let's determine if 1/nYn is approximately normal.
To apply the Central Limit Theorem, we need Yn to be a sum of a large number of independent and identically distributed random variables. However, Yn is not a sum of identical random variables; it is a sum of random variables with alternating signs. Therefore, we cannot directly apply the Central Limit Theorem.
The correct answer is (i) No.
Moving on to find P(Y100 ≥ 20), we can approximate this probability using the properties of the normal distribution.
Since Y100 has a mean of 0 (from the previous calculations) and a variance of var(Y100) = 4 * 100 = 400, we can assume that Y100 approximately follows a normal distribution with mean 0 and variance 400.
To find P(Y100 ≥ 20), we need to calculate the cumulative distribution function (CDF) of the standard normal distribution for the value z = (20 - 0) / sqrt(400).
Using a standard normal distribution table or statistical software, we can find the probability P(Z ≥ z) associated with the z-score.
After finding the corresponding probability from the table or software, we have our approximate value for P(Y100 ≥ 20).
The variance of Yn is 4n, which was calculated earlier to be var(Yn) = 4n.
The variance of Y100, in particular, is var(Y100) = 4 * 100 = 400.
Therefore, the variance of Yn is 400.