How does variance change as sample size increases

Question

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?

Answer 1

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.

Answer 2

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.