Let X and Y be two independent Poisson random variables, with means λ1 and λ2, respectively. Then, X+Y is a Poisson random variable with mean λ1+λ2. Arguing in a similar way, a Poisson random variable X with parameter t, where t is a positive integer, can be thought of as sum of t independent Poisson random variables X1,X2,…,Xt, each of which has mean 1.
Using the information above, and an appropriate limit theorem, evaluate the following limit:
limn→∞∑k>n+n∞e−nnkk!.
To evaluate the limit of the given expression, we can use the limit theorem known as the Poisson Limit Theorem. This theorem states that as the parameter of a Poisson distribution approaches infinity, the distribution converges to a normal distribution with the same mean.
The given expression involves a sum of terms, each of which can be written as a Poisson probability. The term "e^(-n) * (n^n) / k!" corresponds to the probability mass function of the Poisson distribution with parameter n, evaluated at k.
To simplify the expression, let's split the sum into two parts:
1. The terms where k ≤ n.
2. The terms where k > n.
For the terms where k ≤ n, we have n∑k=ne^(-n) * (n^n) / k!
Since there are only finitely many terms in this sum, as n approaches infinity, we can ignore them.
Now let's focus on the terms where k > n. Using the Poisson Limit Theorem, we can approximate each term as follows:
e^(-n) * (n^n) / k! ≈ e^(-n) * (n^n) / sqrt(2πk) * (k/e)^k (approximation using Stirling's approximation for factorials)
By rearranging the terms, we can rewrite this approximation as:
e^(-n)^(k-n) * (n^n) / sqrt(2πk) * (k/n)^k
Now, let's substitute this approximation back into the original expression and evaluate the limit:
lim(n→∞) ∑(k=n+1)^(∞) e^(-n)^(k-n) * (n^n) / sqrt(2πk) * (k/n)^k
Since we are dealing with a sum and a limit, it's not straightforward to evaluate the expression directly. Instead, we can recognize that the sum resembles a Riemann sum for an integral. By taking the limit as n approaches infinity, this Riemann sum converges to an integral.
Approximating the sum as an integral, we have:
lim(n→∞) ∑(k=n+1)^(∞) e^(-n)^(k-n) * (n^n) / sqrt(2πk) * (k/n)^k
≈ ∫(k=n+1)^(∞) e^(-n)^(k-n) * (n^n) / sqrt(2πk) * (k/n)^k dk
Simplifying this integral expression, we can factor out the constants:
≈ (n^n) * e^(-n)^(k-n) / sqrt(2πn) * (1/n)^k * ∫(k=n+1)^(∞) (k/n)^k / sqrt(k) dk
Now, let's focus on the integral part. As n approaches infinity, we can approximate the integral using another limit theorem, the Central Limit Theorem.
The Central Limit Theorem states that the sum of a large number of independent and identically distributed random variables with finite mean and variance will be approximately normally distributed.
In our case, the integrand (k/n)^k / sqrt(k) can be thought of as the probability density function of a random variable. As n approaches infinity, the variable k/n approaches a constant, let's call it x.
Approximating the integral using the Central Limit Theorem, we can rewrite it as follows:
≈ ∫(k=n+1)^(∞) (k/n)^k / sqrt(k) dk
≈ ∫(x=0)^(∞) x^x / sqrt(x) dx
This integral is known as the Frullani integral and its value can be found to be 1.
Therefore, the final evaluation of the limit is:
lim(n→∞) ∑(k=n+1)^(∞) e^(-n)^(k-n) * (n^n) / sqrt(2πk) * (k/n)^k
≈ (n^n) / sqrt(2πn)
Simplifying further, we get:
lim(n→∞) (n^n) / sqrt(2πn) = √(2πn)
= ∞
Thus, the limit of the given expression as n approaches infinity is infinity.