# Proving linearity of variance for independent random variables by induction

Exercise 4.5.9 of Jeff Rosenthal's "A first look at rigorous probability" asks whether or not we can prove the linearity of variance for independent random variables by induction; in the text it is proved by noting that in general

$$Var(\sum_{i=1}^n X_i) = \sum_{i=1}^n Var(X_i) + 2 \sum_{i < j} Cov(X_i, X_j).$$

With independent random variables the covariance terms vanish and we get the desired formula.

In particular he asks the following. If we know that $Var(X + Y) = Var(X) + Var(Y)$ when $X$ and $Y$ are independent, then can we conclude that $Var(X + Y + Z) = Var(X) + Var(Y) + Var(Z)$ when $X, Y$, and $Z$ are independent.

I think the answer is yes. We proved earlier in the text that under these assumptions $X + Y$ and $Z$ are independent. Thus

$$Var(X + Y + Z) = Var((X + Y) + Z) = Var(X + Y) + Var(Z) = Var(X) + Var(Y) + Var(Z).$$

Is this correct?

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## 1 Answer

This is correct, provided you have a convincing proof that if $(X_1,X_2,\ldots,X_{n})$ is independent, for any $n\geqslant3$,, then $(X_1+X_2+\cdots+X_{n-1},X_{n})$ is independent. The fact that if $(X_1,X_2,X_3)$ is independent, then $(X_1+X_2,X_3)$ is independent, is not sufficient to perform the induction.

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