Why does not variance add in general?

Question

Can someone explain why does not variance add in general? Like this: $$Var(X+Y) ≠ Var(X)+Var(Y)$$

Answer 1

Because the formula for variance does not let you add it. While Expectation ($E[X]$) is linear, this does not mean that variance is.

$$ Var(X) = E[X^2] - E[X]^2 $$ which is clearly not linear, since

$$ Var(X + Y) = E[(X + Y)^2] - (E[X + Y])^2 \\ = E[X^2 + 2XY + Y^2] - (E[X] + E[Y])^2 \\ = E[X^2] + 2E[XY] + E[Y^2] - E[X]^2 - E[Y]^2 - 2E[X]E[Y] $$

while

$$ Var(X) + Var(Y) = E[X^2]- E[X]^2 + E[Y^2] - E[Y]^2 $$

Answer 2

\begin{align} Var(X+Y) &= Cov(X+Y, X+Y) \\ &= Cov(X,X)+Cov(X,Y)+Cov(Y,X)+Cov(Y,Y) \\ &= Var(X)+ Var(Y)+2 Cov(X,Y) \end{align}

In general, convariance between $X$ and $Y$ need not be $0$.

Answer 3

The general formula is $$\text{Var}(X+Y) = \text{Var}(X) + \text{Var}(Y) + 2 \text{Cov}(X,Y),$$ so the equality in your question fails precisely when $\text{Cov}(X,Y) \ne 0$.

Answer 4

As a simple counterexample, take $Y=-X$.

Then

$$Var(X-X) \ne Var(X)+Var(-X).$$

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The property holds only when the variables are uncorrelated.

Answer 5

Roughly speaking, the variance represents the second moment of a PDF, this means that its behavior is nonlinear (quadratic) on the variables, that is why

$$ E(X + Y) = E(X) + E(Y) $$

but

$$ {\rm Var}(X + Y) = {\rm Var}(X) + {\rm Var}(Y) + 2{\rm Cov}(X, Y) \not= {\rm Var}(X) + {\rm Var}(Y) $$