I have two independent variables $X$ and $Y$.
$W=X-Y$ when $X\sim \mbox{Bernoulli}\left(1/2\right)$ and $Y\sim N(0,1)$. This puts $\operatorname{Var}(x)=1/4$ and $\operatorname{Var}(Y)=1$, but I have to be misunderstanding something because if $\operatorname{Var}(W)=\operatorname{Var}(X)-\operatorname{Var}(Y)$ then the $\operatorname{Var}(W)$ is negative, which makes no sense.
What am I missing?
$\endgroup$ 13 Answers
$\begingroup$The thing you are misunderstanding is that $\operatorname{Var}(X-Y)\not=\operatorname{Var}(X)-\operatorname{Var}(Y)$.
The accurate formula is $\operatorname{Var}(W)=\operatorname{Var}(X-Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)$.
If $U$ and $V$ are independent, then $\text{Var}(aU+bV)=a^2\text{Var}(U)+b^2\text{Var}(V)$. In our case, the variance of $X-Y$, that is, of $(1)X+(-1)Y$, is the variance of $X$ plus the variance of $Y$.
$\endgroup$ $\begingroup$Variance is not a linear mapping. When $X$ and $Y$ are independent (or more generally uncorrelated), $\operatorname{Var}(X+Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)$. Here you should apply this noticing that $X-Y=X+(-Y)$ and $\operatorname{Var}(-Y)=\operatorname{Var}(Y)$.
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