Possible Rule of thumb for sums of variances?

Mathematics Asked by PortMadeleineCrumpet on January 5, 2022

For questions dealing with insurance problems, I will often run into questions that involve finding aggregate variance. In a certain problem, I’m told that there are 50 policies, and various different amounts of claims with their own various probabilities for said claims. Once I’ve found the expected claim amount per 1 policy, the solution says that I should simply multiply the variance I’ve found by 50. Why is this the case, and not that if $$S=50X$$ then $$Var[S]=Var[50X]=50^2Var[X]$$?

The sum of independent r.v.'s is not $$50$$ times one of them. So the variance properties are that if $$X_1,ldots,X_n$$ are independent (or pairwise uncorrelated) then $$text{Var}(X_1+ldots+X_n)=text{Var}(X_1)+ldots+text{Var}(X_n).$$ So for $$50$$ identically distributed and independent summands it will be $$50text{Var}(X_1)$$.

Answered by NCh on January 5, 2022

Multiplying by 50 the variance of one variabile or summing the same variance of 50 variabiles is not the same thing.

$$V[50X]=50^2V[X]$$

But if $$X_i stackrel{d}=X_j$$

(All the random variabiles have the same distribution and thus also the same variance, say $$V[X]$$)

$$V[X_1+X_2+...+X_{50}]=50V[X]$$

This, IF THE RV'S ARE UNCORRELATED. Otherwise you must sum all the covariances.

$$V[sum_i X_i]=sum_i sum_j Cov(X_i,X_j)$$

These results are very easy to be proved and you can find the proofs in any elementary statistic textbook

Answered by tommik on January 5, 2022