A simple variance question

Say I am paid 1 million dollars per pip on a six sided dice and I roll the dice one time. Now compare this to a game in which I get 1 dollar per pip on a dice, but I roll the die one million times. It seems intuitive to me that a risk averse person would pick the second game, but don't the variances just add because the die rolls are independent leading to the same variance as in the first game? Why is there a difference in variance?

• The variances in the second game add, but the variances in the first game don't (you can think of it as a sum of a million identical (not independent!) dice rolls); if $\sigma^2$ is the variance of a single die then the variance of the second game is $10^6 \sigma^2$ but the variance of the first game is $10^{12} \sigma^2$. – Qiaochu Yuan Nov 24 '17 at 2:39

The standard deviation of the first is $$\sqrt{\frac{35}{12}} 10^6.$$ The standard deviation of the second is $$\sqrt{\frac{35}{12}} 10^3.$$ Variances add, not standard deviations, and the variance of a constant multiple of a random variable is the constant multiple squared times the variance of the original random variable.

Your intuition is correct that you are more likely to end up near $3.5$ million in the second case and less likely to get $1$ million or $6$ million.

Let $(X_k)_{k\in\{0,1...1000\}}$ be a sequence of results for $1001$ independent die rolls.

The variance for the first gamble is: $\mathsf {Var}(1000 X_0) ~{= 10^6\mathsf {Var}(X_0)\\ = 10^6(6^2-1)/12}$

The variance for the second gamble is: $\mathsf {Var}(\sum_{k=1}^{1000} X_k) ~{= \sum_{k=1}^{1000}\mathsf {Var}(X_k) \\= 10^3 \mathsf {Var}(X_1) \\ = 10^3(6^2-1)/12 }$

The former is the variance of the sum of 1000 dependent random variables (in fact the exact same variable), whilst the second is the sum of 1000 independent and identically distributed random variables.

Variance is not additive; covariance in fact is bilinearly additive. Consider the simple case for two random variables. Bilinearity of covariance means that:

$$\mathsf {Var}(A+B)~{= \mathsf {Cov}(A+B,A+B) \\ = \mathsf {Cov}(A,A)+\mathsf {Cov}(A,B)+\mathsf {Cov}(B,A)+\mathsf {Cov}(B,B) \\ = \mathsf {Var}(A)+2\mathsf {Cov}(A,B)+\mathsf {Var}(B)}$$

So in the style of gamble 1, both these random variables are in fact the same variable:

$$\mathsf {Var}(2X_0) ~{=~\mathsf {Cov}(X_0+X_0,X_0+X_0) \\ = \mathsf {Var}(X_0)+2\mathsf {Cov}(X_0,X_0)+\mathsf {Var}(X_0) \\ = 4 \mathsf{Var}(X_0)}$$

However, in the style of gamble 2, both random variables are merely identically distributed, but are independent.   Independence entails that the distinct variables are uncorrolated.

$$\mathsf {Var}(X_1+X_2) ~{=~\mathsf {Cov}(X_1+X_2,X_1+X_2) \\ = \mathsf {Var}(X_1)+2\mathsf {Cov}(X_1,X_2)+\mathsf {Var}(X_2) \\ = \mathsf{Var}(X_1)+\mathsf{Var}(X_2) \\ = 2\mathsf {Var}(X_1) }$$

In intuitive terms: Multiplying the result of the roll of one die by a large number is going to magnify the effect of the single result.   Where as the cummulated results of a multitude of die rolls is going to group the result tighter around the average.