Let X_ n be i.i.d. random variables with mean zero and variance \sigma ^2. Let S_ n=X_1+\cdots +X_ n. Let \Phi stand for the standard normal CDF. According to the central limit theorem, and as n\to \infty, \mathbf{P}(S_ n\leq 2\sigma \sqrt{n}) converges to \Phi (a), where:
a=\,
Furthermore,
\mathbf{P}(S_ n\leq 0) converges to:
(Here, enter the numerical value of the probability.)
According to the central limit theorem, as n approaches infinity, the distribution of the random variable S_n approaches a normal distribution with mean zero and variance n*sigma^2.
Therefore, we can standardize the random variable S_n as follows:
Z_n = (S_n - 0) / sqrt(n*sigma^2) = S_n / sqrt(n*sigma^2).
We want to find the probability that S_n is less than or equal to 2*sigma*sqrt(n), which can be written as:
P(S_n <= 2*sigma*sqrt(n)) = P(S_n / sqrt(n*sigma^2) <= 2*sigma*sqrt(n) / sqrt(n*sigma^2))
= P(Z_n <= 2*sigma / sqrt(n*sigma^2))
= P(Z_n <= 2 / sqrt(n)).
As n approaches infinity, 2 / sqrt(n) approaches 0. Therefore, as n approaches infinity, P(Z_n <= 2 / sqrt(n)) approaches Phi(0), which is equal to 0.5.
So, as n approaches infinity, P(S_n <= 2*sigma*sqrt(n)) converges to Phi(0), where a = 0.
As for the probability P(S_n <= 0), since the random variables X_n have mean zero, the sum S_n will also have mean zero. Therefore, as n approaches infinity, P(S_n <= 0) approaches Phi(0), which is equal to 0.5.
So, as n approaches infinity, P(S_n <= 0) converges to 0.5.
To find the value of 'a', we'll use the Central Limit Theorem (CLT) which states that the sum of a large number of independent and identically distributed random variables will be approximately normally distributed.
In this case, we have S_n = X_1 + X_2 + ... + X_n, where X_n are i.i.d. random variables with mean zero and variance σ^2.
According to the CLT, as n approaches infinity, the standardized variable (S_n - E[S_n]) / sqrt(Var[S_n]) converges in distribution to a standard normal random variable. Here, E[S_n] is the mean of S_n and Var[S_n] is the variance of S_n.
The mean of S_n is given by E[S_n] = E[X_1 + X_2 + ... + X_n] = n * E[X_n] = 0 (since the X_n have mean zero).
The variance of S_n is given by Var[S_n] = Var[X_1 + X_2 + ... + X_n] = n * Var[X_n] = n * σ^2.
So, the standardized variable becomes (S_n - 0) / sqrt(n * σ^2) = S_n / (σ * sqrt(n)).
Now, we need to find the probability that S_n is less than or equal to 2σ * sqrt(n), which can be written as:
P(S_n ≤ 2σ * sqrt(n)) = P(S_n / (σ * sqrt(n)) ≤ 2)
As n approaches infinity, the left-hand side of the inequality converges in distribution to a standard normal random variable, and the right-hand side remains fixed at 2.
Therefore, as n → ∞, P(S_n ≤ 2σ * sqrt(n)) converges to the probability that a standard normal random variable is less than or equal to 2, which is the value of the standard normal cumulative distribution function (CDF) evaluated at 2.
So, a = Φ(2), where Φ represents the standard normal CDF.