Why is the variance of $X-Y$ equal to the sum of the variances when $X,Y$ are independent? 
I have one question about this. I know that if we have $\mathrm{X}_1,\mathrm{X}_2,\ldots,\mathrm{X}_n$ independent and normally distributed random variables, then the sum $\mathrm{X}_1+\mathrm{X}_2+\ldots+\mathrm{X}_n$ has the normal distribution with mean $M_1+M_2+..+M_n$ and variance $\sigma^2_1 + \ldots + \sigma^2_n$.
Why is in this problem the difference $W-M$ the mean obtained by subtraction and variance obtained by addition? Thank you. 
 A: If $X$ and $Y$ are independent random variables, then so are $X$ and $Z$ independent random variables where $Z = -Y$.  Now, 
$$\text{var}(Z) = \text{var}(-Y) = (-1)^2\text{var}(Y) = \text{var}(Y)$$
and so
$$\text{var}(X-Y) = \text{var}(X + (-Y)) = \text{var}(X+Z) = \text{var}(X) + \text{var}(Z) = \text{var}(X) + \text{var}(Y)$$
with nary an explicit mention of the word covariance.
A: Let $X,Y$ be random variables with variances $\sigma^{2}_{x}$ and $\sigma^{2}_{y}$, respectively. It is a fact that ${\rm var}(Z) = {\rm cov}(Z,Z)$ for any random variable $Z$. This can be checked using the definition of covariance and variance. So, the variance of $X-Y$ is 
$$ {\rm cov}(X-Y,X-Y) = {\rm cov}(X,X)+{\rm cov}(Y,Y)-2\cdot{\rm cov}(X,Y) $$
which follows from bilinearity of covariance. Therefore, 
$$ {\rm var}(X-Y) = \sigma^{2}_{x} + \sigma^{2}_{y} - 2\cdot{\rm cov}(X,Y) $$
when $X,Y$ are independent the covariance is 0 so this simplifies to $\sigma^{2}_{x} + \sigma^{2}_{y}$. So, the variance of the difference of two independent variables is the sum of the variances. 
