We know that if $X\sim \mathrm{Pois}(\lambda)$ and $Y\sim \mathrm{Pois}(\mu)$ are independent then $X+Y\sim \mathrm{Pois}(\lambda+\mu).$

This means that if $X_{1},\ldots,X_{n}$ are independent with $X_i \sim \mathrm{Pois}(1/n)$ (and so $E(X_i) = \operatorname{Var}(X_i)=1/n)$ then $S_n=\sum\limits_{i=1}^n X_i\sim \mathrm{Pois}(1)$ for all $n$.

If the CLT holds, it follows that $S_n\sim \mathrm{Pois}(1)$ is a normal distribution, which is not true.

What is wrong with this argument?


3 Answers 3


When you look at the definition of $S_3$, what do $X_1, X_2, X_3$ mean? They're three RV distributed $\mathrm{Pois}(1/3)$.

When you look at the definition of $S_4$, what does $X_2$ mean? IT means a RV $\mathrm{Pois}(1/4)$.

This when you look at the sequence $S_n$, the meaning of $X_i$ keeps changing. The CLT says nothing about things like that. In the CLT, you start with an infinite sequence of RVs $X_i$, and compute their partial sums, etc.

  • $\begingroup$ ...And divide their sum by n, which is not so here. $\endgroup$
    – Did
    Sep 25, 2015 at 12:41
  • $\begingroup$ Good point, Did...thanks! $\endgroup$ Sep 25, 2015 at 15:46

This is an interesting question. The central limit theorem says that the average will approach a normal distribution, meaning that for some measure of closeness to a normal distribution there is some $n$ for which the average of at least $n$ of these variables is at least that close. However, $n$ depends on $\lambda$, and this is the error that you're making by letting $\lambda$ depend on $n$.

To be explicit, CLT follows from approximating the exponent of the moment generating function with a Taylor series. The MGF of Poisson($\lambda$) is $\exp\left(\lambda(e^t-1)\right)$. The exponent can be approximated by $f(t)=\lambda\left(t+\frac{1}{2}t^2+\frac{1}{3}t^3+\cdots\right)$.

It follows from properties of MGFs that the MGF of the average of $n$ of these is $nf\left(\frac{1}{n}\right)=\lambda t+\frac{\lambda}{2n}t^2+\frac{\lambda}{3n^2}t^3+\cdots$, and the high orders of $t$ vanish much faster than the linear and squared terms.

However, if you plug in $\lambda=\frac{1}{n}$, you would get that the average approaches something not normal, and as a result you should expect the sum of the averages to be normal either.


if $X_1,\ldots,X_n$ are independent with $X_i \sim \mathrm{Pois}(1/n)$ (and so $E(X_i) = \operatorname{Var}(X_i)=1/n)$ then $S_n=\sum\limits_{i=1}^n X_i\sim \mathrm{Pois}(1)$ for all $n$.

If the CLT holds, it follows that $S_n\sim \mathrm{Pois}(1)$ is a normal distribution

No, you're misapplying the theorem.

Suppose $n=6$. Then you have $X_1,\ldots,X_6 \sim \mathrm{i.i.d. Poisson}\left(\dfrac 1 6 \right)$.

Then $X_1+\cdots+X_6 \sim \mathrm{Poisson}(1)$. And $X_1+\cdots+X_{600} \sim \mathrm{Poisson}(100) \ne \mathrm{Poisson}(1)$. And $\mathrm{Poisson}(100)$ is approximately normal with expectation $100$ and variance $100$ (so the standard deviation is $10$). But use a continuity correction here: if you want $\Pr(X_1+\cdots+X_{600}>95)$, realize that that is the same as $\Pr(X_1+\cdots+X_{600}\ge96)$ and so plug in $95.5$.

If you then change the distribution of $X_1,\ldots,X_{600}$ to $\mathrm{Poisson}\left(\dfrac 1 {600}\right)$, then their sum is indeed distributed as $\mathrm{Poisson}(1)$, which is not so close to normal. But if $X_1,\ldots,X_{6000000}\sim\mathrm{i.i.d. Poisson}\left(\dfrac 1 {600}\right)$ then again you get something that is approximately normal.

The central limit theorem says if $X_1,X_2,X_3,\ldots\sim\mathrm{i.i.d.}$ with a finite variance, then the distribution of $$ \frac{(X_1+\cdots+X_n)-\operatorname{E}(X_1+\cdots+X_n)}{\operatorname{SD}(X_1+\cdots+X_n)} $$ approaches the standard normal distribution as $n\to\infty$. It does not say that if you keep changing the distribution of each of the random variables as $n$ increases, then the same thing happens.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .