# Is the distribution of sample mean of Bernoulli random variable a Binomial distribution?

As written in the title, is the distribution of sample mean of Bernoulli random variable a Binomial distribution? And I was taught that we can approximate Binomial distribution to normal distribution due to CLT. Is the title the reason we can approximate it?

I think, if $$X \sim Bern(p)$$, and say $$X_n$$ is the sum of the elements of the sample of scale $$n$$, i.e. $$X_n=n\bar{X_n}$$ then $$Pr(X_n=k)={n \choose k}p^kq^{n-k}$$so $$X_n \sim Bin(n,p)$$.

And due to CLT, which ststes $$\sqrt{n}( \bar{X_n} -\mu)\overset {d}{\to} N(0,\sigma^2),$$ the distribution of sample mean $$\bar{X_n}$$ tents to follow $$N(\mu,\frac{\sigma^2}{n})$$ as $$n$$ gets bigger, and therefore $$X_n=n\bar{X_n}$$ tends to follow $$N(n\mu, n\sigma^2)$$.

And if we apply this consequence to random variable $$X \sim Bern(p),$$ then since $$E(X)=p$$ and $$V(x)=pq$$,
$$X_n \sim Bin(n,p)$$ tends to follow $$N(n\mu ,n \sigma^2 )=N(np, npq).$$

Is this the right way to suggest we can approximate binomial distribution to normal distribution by CLT? Or is it wrong and does there exist other way to suggest it?

#Since I'm new to statistic and I'm Korean, I would appreciate if you forgive some improper expressions.

• I think you should write $P(X_n = k) = \binom{n}{k}p^kq^{\color{red}{n-k}}$. Commented Dec 26, 2023 at 11:30
• @Riemann Oh that was right. I will edit Commented Dec 26, 2023 at 11:49

## 1 Answer

No, as you said, sum of n independent bernulli random variables is binomial but if $$Y\sim Bin(n,p)$$ then $$aY$$ is not binomial (a≠1). You can compare the support of $$aY$$ to binomial to see why $$aY$$ is not binomial. The support of binomial is like $$\{0,1,...,n\}$$ but support of $$aY$$ is $$\{0,a,2a,...,an\}$$ which is different. Also there are other ways to show that $$aY$$ is not binomial but if you plot the probability distributions of $$aY$$ and $$Y$$, you see that they are similar but it doesn't mean that $$aY$$ is binomial.

Also, take a look at this question.