I am working on a problem from probability theory and am a little bit stuck.
I know that the formula for Var(X + Y) is Var(X) + Var(Y) + 2Cov(X,Y)
Does this mean that for Var(X - Y) it is just:
$\text{Var}(X) - \text{Var}(Y) - 2\text{Cov}(X,Y)$?
$\endgroup$ 52 Answers
$\begingroup$It will be $\text{Var}(X) + \text{Var}(Y) - 2\text{Cov}(X,Y)$, because $\text{Var}(-Y) = \text{Var}(Y)$.
$\endgroup$ 2 $\begingroup$$Var (X) = E[X^2] - E[X]^2$
The definition of variance.
$Var (X-Y) = $$E[(X-Y)^2] - E[X-Y]^2\\ E[X^2 - 2XY + Y^2] - E[X-Y]^2$
Linearity of expectation:
$E[X^2 - 2XY + Y^2] = E[X^2] + E[Y^2] - 2E[XY]$ and $E[X-Y] = E[X] - E[Y]$
$Var (X-Y) = $$E[X^2] - 2E[XY] + E[Y^2] - (E[X]^2 - 2E[X]E[Y] + E[Y]^2)\\ E[X^2] - E[X]^2 + E[Y^2] - E[Y]^2 - 2(E[XY] - E[X]E[Y])$
Now note that:\begin{align} Cov(x,y)&=E[(x-E[x])](y-E[y])]\\ &=E[xy]-E[x E[y]]-E[y[E[x]]+ E[x]E[y]\\ &=E[xy]-E[x]E[y]-E[y]E[x]+E[x]E[y]\\ &=E[xy]-E[x]E[y] \end{align}
Which immediately gives the result desired in terms of the covariance.
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