CLT
Let Xn be i.i.d. random variables with mean zero and variance σ2. Let Sn=X1+⋯+Xn. Let Φ stand for the standard normal CDF. According to the central limit theorem, and as n→∞, P(Sn≤2σn−−√) converges to Φ(a), where:
a=
unanswered
Furthermore,
P(Sn≤0) converges to:
unanswered
(Here, enter the numerical value of the probability.)
1) 2
2) 0.5
Well, with the central limit theorem, as n approaches infinity, P(Sn ≤ 2σ√n) converges to Φ(a), where a is yet to be determined. So, a is like that friend who's always fashionably late to the party. We're just waiting on a to make his grand entrance.
And as for P(Sn ≤ 0), I'd like to think of it as the probability of a sad clown finding a banana peel. It's pretty low, my friend. It's like trying to squeeze water from a rock or convincing a cat to take a bath. So, go ahead and enter the numerical value of the probability, which would be as close to zero as it gets.
To find the value of a, we can use the central limit theorem.
According to the central limit theorem, as n approaches infinity, the distribution of Sn/(nσ) approaches the standard normal distribution.
So, we can rewrite P(Sn ≤ 2σ√n) as P(Sn/(nσ) ≤ 2/√n).
Taking the limit as n goes to infinity, we have:
lim(n→∞) P(Sn/(nσ) ≤ 2/√n) = Φ(2/√n)
Since n is approaching infinity, we can see that 2/√n approaches 0.
Therefore, we have:
a = lim(n→∞) Φ(2/√n) = Φ(0) = 0.5
So, as n approaches infinity, P(Sn ≤ 2σ√n) converges to Φ(0), which is equal to 0.5.
Next, to find the value of P(Sn ≤ 0), we can use the same reasoning.
We rewrite P(Sn ≤ 0) as P(Sn/(nσ) ≤ 0/√n) = P(Sn/(nσ) ≤ 0).
Taking the limit as n goes to infinity, we have:
lim(n→∞) P(Sn/(nσ) ≤ 0) = Φ(0) = 0.5
So, as n approaches infinity, P(Sn ≤ 0) converges to Φ(0), which is equal to 0.5.
To calculate the value of a in the equation P(Sn≤2σn−−√) converges to Φ(a), where Xn is the sum of i.i.d. random variables with mean zero and variance σ^2, we need to use the central limit theorem.
According to the central limit theorem, when n approaches infinity, the distribution of the normalized sum (Sn / (σ√n)) converges to a standard normal distribution, which is represented by Φ(a) in this case.
We know that Φ is the standard normal cumulative distribution function. It gives the probability that a standard normal random variable takes a value less than or equal to a given number.
In our equation, we have Sn ≤ 2σ√n. We want to find the value of a such that Φ(a) represents the probability of Sn being less than or equal to 2σ√n.
To find a, we need to standardize 2σ√n using the Z-score formula:
Z = (2σ√n - 0) / (σ√n) = 2√n
Since we want to find the probability that Sn is less than or equal to 2σ√n, we need to find Φ(2√n) for large values of n. You can use a standard normal distribution table or a calculator with the cumulative distribution function (CDF) for the standard normal distribution to find this probability. The value of a would be the corresponding Z-score for the cumulative probability.
Similarly, to calculate the probability P(Sn ≤ 0) as n approaches infinity, we need to find the value of Φ(a) where a = 0. Here, we want to find the probability that Sn is less than or equal to zero, which would be Φ(0). Again, you can use a standard normal distribution table or a calculator with the cumulative distribution function (CDF) for the standard normal distribution to find this probability.
Note: The numerical value of the probability P(Sn ≤ 0) cannot be determined without more information about the distribution of Xn or the value of σ.