# Why is the variance of X−YX-Y equal to the sum of the variances when X,YX,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.

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
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.