# How does variance change as sample size increases

## Situation:

n possibilities each have their own probability of happening, and their own payout when they do.

So expected payout $E_n$ is $\sum\limits_{i=1}^n \text{probability}_i*\text{payout}_i$

And the expected variance is $\sum\limits_{i=1}^n ( \text{payout}_i- \text{expected payout})^2* \text{probability}_i$

## Question:

If this activity is repeated 5 times, the expected payout is $5* \text{expected payout}$

But the variance is not 5 times the expected variance; it should get proportionally smaller as this activity goes on. But how do I calculate the variance through iterations?

The variance of a sum is the sum of the variances plus twice the sum of the covariances.

e.g. $$\,\, \operatorname{Var}(X+Y+Z)$$ $$= \operatorname{Var}(X)+\operatorname{Var}(Y)+\operatorname{Var}(Z) \\+ 2 [\operatorname{Cov}(X,Y)+ \operatorname{Cov}(X,Z) + \operatorname{Cov}(Y,Z)]$$

If your events are independent, those covariances are zero, and that reduces to "the variance of a sum is the sum of the variances":

$$\operatorname{Var}(X+Y+Z)= \operatorname{Var}(X)+\operatorname{Var}(Y)+\operatorname{Var}(Z)$$

So if you repeat it 5 times, yes, the variance of the total is indeed 5 times larger.

The standard deviation therefore increases as the square root of the number of repetitions, which may be what you're anticipating.

If you are interested in the sum of payouts then, as @Glen_b suggests, variance gets larger with more repetitions of the experiment: $kVar_n$ after $k$ repetitions.

If you are interested in the average payout, then variance is $\frac{Var_n}{k}$ after $k$ repetitions. Since you say:

[variance] should get proportionally smaller as this activity goes on.

I guess you are interested in the average. This also assumes independence.