Let X1,X2,… be a sequence of independent random variables, uniformly distributed on [0,1] . Define Nn to be the smallest k such that X1+X2+⋯+Xk exceeds cn=n2+n12−−−√ , namely,
Nn = min{k≥1:X1+X2+⋯+Xk>cn}
Does the limit
limn→∞P(Nn>n)
exist? If yes, enter its numerical value. If not, enter −999 .
correction:
cn = (n/2) + sqrt(n/12)
To determine if the limit of limn→∞P(Nn>n) exists, we need to consider the properties of the sequence of random variables Nn.
First, let's notice that Nn is a non-negative integer random variable. It represents the number of terms needed to exceed the threshold cn in the sequence X1, X2, ..., Xk.
Next, we can rewrite the probability expression as P(Nn > n) = 1 - P(Nn ≤ n). In other words, it is the probability that Nn is greater than n.
Now, let's break down the calculation of P(Nn ≤ n). The event Nn ≤ n occurs when the sum X1+X2+⋯+Xk does not exceed cn for all k up to n. Since X1, X2, ..., Xk are independent and uniformly distributed on [0, 1], their sum follows the Irwin-Hall distribution with parameters n. This distribution represents the sum of n independent uniform(0, 1) random variables.
Using the Irwin-Hall distribution, we can calculate the probability P(Nn ≤ n) as the probability that the sum of the first n terms does not exceed cn. This can be expressed as:
P(Nn ≤ n) = P(X1 + X2 + ⋯ + Xn ≤ cn)
To proceed further, we need to compute the cumulative distribution function (CDF) of the Irwin-Hall distribution with parameters n. The CDF gives us the probability that the sum of n independent uniform(0, 1) random variables is less than or equal to a given value.
Let's denote the CDF of the Irwin-Hall distribution with parameters n as Fn(x). We can then substitute cn for x in the CDF:
P(Nn ≤ n) = Fn(cn)
Finally, we can calculate the desired limit as:
limn→∞P(Nn > n) = limn→∞(1 - Fn(cn))
To find the numerical value of the limit, we need to evaluate the CDF Fn(x) and take the limit as n approaches infinity.
Unfortunately, without the specific value of n and a given formula for the Irwin-Hall CDF, it is not possible to provide a numerical value for the limit.