Prove the central limit theorem for a sequence of i.i.d. Bernoulli($p$) random variables, where $p\in(0,1)$.

I am trying to do this by computing the moment generating function of the object I want the limit of and use Taylor's expansion to show that it converges to the moment generating function of a standard normal.

Attempt: For a random variable X, we have the moment generating function $$M_{X}(t)=\mathbb{E}[e^{tX}]$$ and if we expand using the Taylor series of $e^{tX}$ we get $$M_X(t)=\sum_{n=0}^\infty \frac{\mathbb{E}[X^n]}{n!}t^n.$$ So, we have in particular $$M_x^{(n)}(0)=\mathbb{E}[X^n].$$

Now for the proof, we have a Bernoulli random variable is 1 with probability $p$ and 0 with probability $(1-p)$. First we want the moment generating function for a Bernoulli random variable. In particular, we have $$M_{Bernoulli(p)}=(1-p)+pe^t=1+(e^t-1)p$$ Then, we have a binomial random variable as the sum of n independent Bernoulli variables. So, $$M_{binomial(n,p)}=((1-p)+pe^t=(1+(e^t-1)p)^n$$ Suppose that $p=\lambda/n$ and notice that $$M_{binomial(n,\lambda/n)}=\bigg(1+\frac{(e^t-1)\lambda}{n}\bigg)^n \rightarrow e^{\lambda(e^t-1)}$$

I am not sure how to finish the proof though or if I am even on the right track...

  • 1
    $\begingroup$ What you have done so far is merely deriving (correctly) Binomial becoming Poisson under the limit $n \to \infty, p \to 0$ with $np = \lambda$ held constant. The statement of Central Limit Theorem involves $\frac{\bar X - \mu}{ \sigma}$. Where's your analysis on $\bar X$? $\endgroup$ – Lee David Chung Lin Feb 2 at 2:05
  • $\begingroup$ Oh okay. Thank you. So I want to find the mean and the std deviation for the random variable next? $\endgroup$ – MathIsHard Feb 2 at 2:48
  • 1
    $\begingroup$ There are many ways to do this. Since you wanted to use the moment generating function, you should find the MGF of $M_Y(t)$ where $Y \equiv \bar X = \frac1n \sum X_i$ and then further consider the MGF of $\sqrt{n}\frac{Y - \mu}{ \sigma}$ $\endgroup$ – Lee David Chung Lin Feb 2 at 3:36
  • $\begingroup$ oh okay. Thank you. I will give that a try. I appreciate your time $\endgroup$ – MathIsHard Feb 2 at 4:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.