Limit of rescaled Binomials

Question

Let Xn be a binomial random variable with parameters n and p = lambda/n, where lambda is a fixed positive number.

Let k be a natural number. As n tends to infinity, the probability mass function P(Xn = k) converges to a number that only depends on lambda and k. What is the limit?

Answer 1

To find the limit of P(Xn = k) as n tends to infinity, we can use the concept of the Poisson distribution.

The probability mass function of a binomial random variable Xn with parameters n and p can be written as:

P(Xn = k) = nCk * p^k * (1-p)^(n-k)

Where nCk represents the binomial coefficient.

Now, let's substitute p = lambda/n into the equation:

P(Xn = k) = nCk * (lambda/n)^k * (1-(lambda/n))^(n-k)

Simplifying the expression:

P(Xn = k) = nCk * (lambda/n)^k * ((n-lambda)/n)^(n-k)

As n tends to infinity, the binomial coefficient nCk can be approximated by the binomial coefficient of the limiting distribution, which is the Poisson distribution with parameter lambda. We can write:

nCk ≈ lambda^k / k!

Also, observe that ((n-lambda)/n)^(n-k) tends to 1 as n tends to infinity.

Substituting these approximations into the equation:

P(Xn = k) ≈ (lambda^k / k!) * (lambda/n)^k * 1

= (lambda^k / k!) * (lambda^k / n^k)

= (lambda^2k / (k!) * n^k)

As n tends to infinity, the term (lambda^2k / (k!) * n^k) tends to 0 as the denominator n^k increases faster than the numerator lambda^2k.

Therefore, the limit of P(Xn = k) as n tends to infinity is 0 for k ≠ 0, and when k = 0, it converges to (lambda^0 / 0!) = 1.

In summary, the limit of the probability mass function P(Xn = k) as n tends to infinity is:
1 for k = 0
0 for k ≠ 0

The limiting distribution is a degenerate Poisson distribution with parameter lambda, where the probability of observing a non-zero value is 0.