Convergence of Moment-Generating Functions (MGF) → Convergence of Cumulative Distribution Functions (CDF)

That is, let 𝑋1, 𝑋2, … be a sequence of random variables, each with:

• $MG{F}_{X_i}(t)$
• $CD{F}_{X_i}(x)$

If both:

• $li{m}_{i\to \infty }{M}_{X_i}(t)=M_X(t)\text{ ∀ }t\text{ in a neighborhood of }0$
• $M_X(t)\text{is an MGF}$

Then:

• there exists a random variable 𝑋 with CDF 𝐹_𝑋(𝑥) where:
  • $li{m}_{i\to \infty }{F}_{X_i}(x)=F_X(x)$
• and this random variable 𝑋 has moments determined by MGF 𝑀_𝑋(𝑡)

Example

Binomial Converges to Poisson

Show that a binomial distribution will converge into a Poisson distribution.

Bin(𝑛,𝑝) can be approximated by Pois(𝜆) using 𝜆=𝑛𝑝.

Let:

• 𝑋 = Bin(𝑛,𝑝)
• 𝑌 ~ Pois(𝑛𝑝)

Then:

• 𝑃(𝑋=𝑥) ≈ 𝑃(𝑌=𝑥) for large 𝑛 and small 𝑝

Let’s show MGFs converge

First a lemma:

If:

• $a_n\underset{n\to \infty }{\to }a$

Then:

• $[1+a_n/n{]}^n\underset{n\to \infty }{\to }{e}^a$

Reminder:

• see Moment-Generating Function - Binomial Distribution
  $M_X(t)=[1-p+p{e}^t{]}^n$
• see Moment-Generating Function - Poisson Distribution
  $M_Y(t)=e^{𝜆(e^t-1)}$

PROOF

• $M_X(t)=[1-p+p{e}^t{]}^n$
• $M_X(t)=[1+p\ast (e^t-1){]}^n$
• $M_X(t)=[1+p\ast (n/n)\ast (e^t-1){]}^n$
• $M_X(t)=[1+pn\ast (e^t-1)/n{]}^n$
• $M_X(t)=[1+𝜆\ast (e^t-1)/n{]}^n$

Now let 𝑎_𝑛 = (𝑒^𝑡 - 1)𝜆

• $M_X(t)=[1+a_n/n{]}^n$

Then as 𝑛 goes to ∞

• $M_X(t)=[1+a_n/n{]}^n\underset{n\to \infty }{\to }{e}^a$

where:

• $a=li{m}_{n\to \infty }(e^t-1)𝜆$

Then:

• $M_X(t)\underset{n\to \infty }{\to }exp(e^t-1)𝜆$

This is 𝑀_𝑌(𝑡) the MGF for Poisson

So:

• $M_X(t)\underset{n\to \infty }{\to }{M}_Y(t)$

Therefore:

• $Bin(n,p)\underset{n\to \infty }{\to }Pois(𝜆)$

where:

• 𝑝 = 𝜆/𝑛