Connection between Gauss and Poisson distribution?

Question

I know that these distributions are connected by the central limit theorem.

But as it is written here: The central limit theorem says that the distribution of the mean of N draws from a probability distribution approaches a Gaussian of width $\frac{\sigma}{\sqrt{N}}$ for large N.

In the same script, in Eq(70) he says the standard deviation of a poisson distribution is $\sigma = \sqrt{\lambda t}$

Below Eq(79) he says the standard deviation of the gaussian (as limit of the poisson) is also $\sigma = \sqrt{\lambda t}$.

For me this is confusing, where is with the $\frac{1}{\sqrt{N}}$ ?

In our case, the central limit theorem says that the distribution of the mean of N draws from the poisson distribution approaches a Gaussian.

As in the example of the radioactive decay:

I measure N times, the number of decays during the time t (this is the mean) and consider the distribution. This should be a Gaussian? But in this script, he do not sample over anything, he just says that he consider m >> 1 with m as the number of events during t. Is this right? I mean it should be right - he is professor in Harvard - but I don't get it.

Answer 1

If you have $N$ independent Poisson random variables, each with parameter $\lambda$, their sum is Poisson with parameter $N\lambda$. So a Poisson random variable with large parameter can be considered as a sum of independent Poisson random variables, and you can apply the Central Limit Theorem to that.

Answer 2

The answer on my question is in the same script:

Sometimes we add the values of the draws from a distribution instead of averaging them. In this case, the mean $\bar{x}$ grows as $N\bar{x}$ and the standard deviation $\sigma$ grows like $\sqrt{N}\sigma$.

Where N is somehow the time t we measure (for example the radioactive decay)