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This question arrises from the lecture of a book called "Data mining" by Witten and Frank. I try to understand a statement and there is one point I just do not understand nor do I find any explanation:

We are considering a Bernoulli process which has a success rate p and thus mean p and variance p(1-p). So far so good. Now the text says

If N trials are taken from a Bernoulli process, the expected success rate f=S/N is a random variable with the same mean p; the variance is reduced by a factor of N to p(1-p)/N.

Now why is that?

If this question is trivial please excuse. I am not a mathematician but a physicist trying to get some knowledge in Data Science.

Thanks in advance!

Btw: I found this question here already but it makes me even more confused: Variance of n Bernoulli Trials

croessler
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1 Answers1

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Variance and expected value are additive. So the expected value of $n$ trials where each has expected value $p$ is $np$, and the variance of $n$ trials where each has variance $p(1-p)$ is $np(1-p)$. Consequently, the standard deviation goes from $\sqrt{p(1-p)}$ to $\sqrt{np(1-p)}$. But in your setting, you do not consider the sum, but the average ("rate") of $n$ Bernoulli trials. Hence we divide the expected value by $n$, arriving back at $p$; and we divide the standard deviation by $n$ - or equivalently, we divide the variance by $n^2$, which gives us a variance of $\frac{np(1-p)}{n^2}=\frac{p(1-p)}n$. Intuitively, the more often we repeat a Bernoulli trial, the less does the observed rate deviate from the expected rate.

Hagen von Eitzen
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