# Binomial experiment verify sampling statistics calculation

Suppose you have a binomial experiment with 1 million trials and chance of success 0.25. At the end of the 1 million trials you record the number of successes. You repeat this experiment 100 million times, independently. What should you estimate the sample mean of succeses and sample second moment of success to be?

I got that the sample mean should still be the mean of an individual trial $n*p = 1,000,000*.25$, or 250,000 while the sample second moment should be $6.25*10^{10}$ it seems like an extremely large number, is it correct? I got the sample second moment by adding the square of the sample mean to the individual variance of each trial. ie: $\sigma^2 + (\bar X )^2$. I got the individual variance $\sigma^2 = n*p*(1-p)$ each experiment as 187500 approximately. Please verify if I have the right numbers.

EDIT: The sample variance of the 100 million experiments is 0.00188.

• What do you mean by sample product moment? Commented Apr 2, 2017 at 15:53
• I mean sample second moment sorry Commented Apr 2, 2017 at 17:12

let $X_i\sim Binom(N,p)$ and n be number of simulations

we know

$E(\bar{X_i})=E(\frac{\sum X_i}{n})=\frac{\sum E(X_i)}{n}=\frac{n*E(X_i)}{n}(because~X_i's~are~identical~for~each~simulation)=Np$

and

$E(\bar{X_i^2})=E(\frac{\sum X_i^2}{n})=\frac{\sum E(X_i^2)}{n}=\frac{n*E(X_i^2)}{n}(because~X_i's~are~identical~for~each~simulation)=(Np)^2+Np(1-p)$

you can see raw moments from sample is same as population raw moments.

so firsts moment is 250,000 and second moment is 62,500,187,500

First - n∗p∗(1−p) is not .00188 - it's much larger (n is a million!). Second - of course the numbers you get are large - a million is a large number, and the variance is the square of the standard deviation and therefore is very large.י

• made some edits: The 0.00188 should be the sample variance right? I basically divided the individual variance $\sigma^2$ by 100 million. The individual variance I got by np(1-p) = 187500. and i still get the same sample second moment $6.25*10^{10}$. Hope this is right! Commented Apr 3, 2017 at 0:37
• by dividing 100 million, you were getting variance of mean Commented Apr 3, 2017 at 13:26
• @HemantRupani yes i mean the variance of the sample mean thanks for clarifying Commented Apr 3, 2017 at 14:39