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Confused on the general addition rule for variances.

Why is it that when two random variables, X and Y, have a perfect positive correlation (p=1) their standard deviations add. But when they are uncorrelated (p=0) their variances add?

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2 Answers 2

The equation $\text{Var}(X+Y) = \text{Var}(X) + \text{Var}(Y) + 2 \text{Cov}(X,Y)$ covers all cases. Note that $ \text{Cov}(X,Y) = \rho\; \sigma_X\; \sigma_Y$.

Where it comes from is the linearity of expected value:

$$\eqalign{E[X+Y] &= E[X] + E[Y]\cr E[(X+Y)^2] &= E[X^2 + 2 X Y + Y^2] = E[X^2] + 2 E[XY] + E[Y^2]\cr}$$

and the definitions of variance and covariance in terms of expected values.

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thanks for the help. so: $[std(x+y) = sqrt[std(x)^2 + std(y)^2 + 2.std(x).std(y)] = sqrt[(std(x)^2 + std(y))^2] = std(x) + std(y)]$ Correct? –  crimsontide Nov 29 '12 at 3:45
    
No, the middle expression (in the case $\rho = 1$) is $\sqrt{(std(x)+std(y))^2}$. –  Robert Israel Nov 29 '12 at 6:33
    
Wow, this has gone a long time without an upvote. Community threw it at the front page. (+1) –  robjohn May 22 '13 at 22:22

Further illumination on Robert Israel's answer:

Starting with the formula $$ \begin{align} \mathrm{Var}(X) &=\mathrm{E}\left((X-\mathrm{E}(X))^2\right)\\ &=\mathrm{E}(X^2)-\mathrm{E}(X)^2\\ \end{align} $$ we get $$ \begin{align} \mathrm{Var}(X+Y) &=\mathrm{E}\left(X^2+2XY+Y^2\right)-\left(\mathrm{E}(X)^2+2\mathrm{E}(X)\mathrm{E}(Y)+\mathrm{E}(y)^2\right)\\ &=\left(\mathrm{E}(X^2)-\mathrm{E}(X)^2\right)+\left(\mathrm{E}(Y^2)-\mathrm{E}(Y)^2\right)+2\Big(\mathrm{E}(XY)-\mathrm{E}(X)\mathrm{E}(Y)\Big)\\[5pt] &=\mathrm{Var}(X)+\mathrm{Var}(Y)+2\mathrm{Cov}(X,Y) \end{align} $$ Furthermore, using Hölder, $$ \begin{align} \left|\mathrm{Cov}(X,Y)\right|^2 &=\left|\mathrm{E}\Big((X-\mathrm{E}(X))(Y-\mathrm{E}(Y))\Big)\right|^2\\ &\le\mathrm{E}\left((X-\mathrm{E}(X))^2\right)\mathrm{E}\left((Y-\mathrm{E}(Y))^2\right)\\ &=\mathrm{Var}(X)\mathrm{Var}(Y) \end{align} $$ For perfect positive correlation, we have $\mathrm{Cov}(X,Y)=\mathrm{Var}(X)^{1/2}\mathrm{Var}(Y)^{1/2}$; therefore, $$ \begin{align} \mathrm{Var}(X+Y) &=\mathrm{Var}(X)+\mathrm{Var}(Y)+2\mathrm{Var}(X)^{1/2}\mathrm{Var}(Y)^{1/2}\\ \mathrm{Std}(X+Y)^2 &=\mathrm{Std}(X)^2+\mathrm{Std}(Y)^2+2\mathrm{Std}(X)\mathrm{Std}(Y)\\ \mathrm{Std}(X+Y) &=\mathrm{Std}(X)+\mathrm{Std}(Y) \end{align} $$

For independence, we have $\mathrm{Cov}(X,Y)=0$; therefore, $$ \mathrm{Var}(X+Y)=\mathrm{Var}(X)+\mathrm{Var}(Y) $$

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