In this problem, you may find it useful to recall the following fact about Poisson random variables. 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!.
Why did the Poisson random variable start a shoe store? Because it wanted to find its sole mate!
Now, let's tackle this limit using our knowledge of Poisson random variables. We see that the sum involves the Poisson random variable with parameter n.
Since we're taking the limit as n approaches infinity, we can use the law of large numbers. This law states that as the number of trials goes to infinity, the sample mean approaches the population mean.
In this case, our sample mean is k/n and the population mean is 1. So, as n approaches infinity, the expected value of k/n approaches 1.
Now, let's simplify the expression inside the limit: e^(-n/n) * (k/n)^k/k!
Notice that e^(-n/n) simplifies to e^(-1). Also, as we take the limit and k/n approaches 1, the term (k/n)^k/k! approaches e^1/1!, which is just 1.
So, we are left with e^(-1) * 1 = e^(-1).
Therefore, the value of the limit is e^(-1), which is approximately 0.3679.
I hope my explanation wasn't too random for you!
To evaluate the given limit, we can express the sum as a sum of Poisson probabilities. Let's break down the steps:
Step 1: Rewrite the given sum in terms of Poisson random variables.
Since it is stated that a Poisson random variable with parameter t can be thought of as a sum of t independent Poisson random variables with mean 1, we can rewrite the sum as follows:
limn→∞ ∑k>n (n/e)^n * (1/n)^k/k!
Step 2: Simplify the expression.
Using properties of exponents and simplifying the fractions, we can simplify the expression further:
limn→∞ ∑k>n [(n/e) * (1/n)^k/k!]^n
Step 3: Apply the Poisson random variable property.
Using the fact stated about Poisson random variables, we can rewrite the expression inside the brackets as a Poisson random variable:
limn→∞ ∑k>n [Poisson random variable with mean λ = (n/e) * (1/n)^k/k!]^n
Step 4: Use the fact that the sum of Poisson random variables with independent means is also Poisson.
Since the terms inside the sum are Poisson random variables with a common mean λ = (n/e) * (1/n)^k/k!, we can rewrite the entire sum as a single Poisson random variable:
limn→∞ Poisson random variable with mean λ = (n/e) * (1/n)^n
Step 5: Simplify the expression further.
Using properties of exponents and rearranging terms, we can simplify the expression inside the limit:
limn→∞ Poisson random variable with mean λ = (n/e)^n / (n/e)
Step 6: Evaluate the limit.
As n approaches infinity, the expression (n/e)^n becomes infinitely large. However, dividing it by (n/e) still yields infinity. Therefore, the limit is:
limn→∞ Poisson random variable with mean λ = ∞
So, the given limit is equal to infinity.
To evaluate the given limit, we can start by analyzing the expression inside the limit:
∑(k > n) n∞ e^(-n/n) k!/n!
The key here is to recognize that k!/n! can be simplified using the property of factorials:
k! = k(k-1)(k-2)...(n+1)(n!),
whereas n! = n(n-1)(n-2)...(1).
Therefore, k!/n! can be rewritten as:
k!/n! = k(k-1)(k-2)...(n+1).
Now, let's rewrite our expression:
∑(k > n) n∞ e^(-1) (k(k-1)(k-2)...(n+1)).
Next, let's consider the sum as a general form of the sum of a Poisson random variable:
∑(k > n) n∞ e^(-1) (k1(k2-1)(k3-2)...(n+1)),
where k1, k2, k3, ..., kn+1 are the independent Poisson random variables.
Since each term in the sum is a product of independent Poisson random variables, we can use the property mentioned at the beginning of the problem: "a Poisson random variable X with parameter t, where t is a positive integer, can be thought of as the sum of t independent Poisson random variables."
Therefore, we can rewrite the sum as:
∑(k > n) n∞ e^(-1) (n-(k-1))/(n+1)!
Notice that the numerator n-(k-1) is the number of terms in the sum, which can be written as n+1-(k-1).
Now, let's simplify the expression further:
∑(k > n) n∞ e^(-1) [(n+1-(k-1))/(n+1)!],
which becomes:
∑(k > n) n∞ e^(-1) [1/(n+1)!] ∑_(i=0)^(n+1-(k-1)) (-1)^i/(n+1-(k-1))!.
The last step before evaluating the limit is to recognize the pattern in the sum:
∑_(i=0)^(n+1-(k-1)) (-1)^i/(n+1-(k-1))!
= (n+1-(k-1))!/(n+1-(k-1))! - (n+1-(k-1)+1)!/(n+1-(k-1))! + ...
The terms cancel out, leaving only the first term:
= 1/(n+1-(k-1))!.
Finally, we have:
∑(k > n) n∞ e^(-1) [1/(n+1)!] * 1/(n+1-(k-1))!
As n approaches infinity, the term 1/(n+1)! approaches zero since the factorial grows much faster than any polynomial expression in n. Thus, we can conclude that the entire expression approaches zero as well.
Therefore, the limit as n approaches infinity of the given expression is zero:
lim_(n→∞) ∑(k > n) n∞ e^(-n/n) k!/n! = 0.