# 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.

• If $M\sim N(68,3^2)$ then $-M\sim N(-68,3^2)$. Therefore $W-M=W+(-M)\sim N(65+(-68),1^2+3^2)$. This is, the values M1,M2,...Mn are not necessarily positive. I hope this helps.
– user10525
Apr 21, 2012 at 14:54
• @Procrastinator I thought of what you wrote a bit and it makes sense. Thank you Apr 21, 2012 at 15:09
• If variances subtracted, then (for uncorrelated random variables) the variance of $X-Y$ would be negative whenever $\sigma_Y^2$ exceeded $\sigma_X^2$. The only time I have seen variances subtract is in the identity $$\operatorname{cov}(X+Y,X-Y) = \operatorname{var}(X) - \operatorname{var}(Y)$$ which applies to all random variables with finite variances, whether correlated or uncorrelated, dependent or independent, normal or abnormal etc. Apr 23, 2012 at 15:46
• @Andrew, this recent question answers this question in greater generality: stats.stackexchange.com/questions/31177/… Jun 29, 2012 at 0:34

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.

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.

• +1 since this only requires the fact that the OP already mentioned in the question: ${\rm var}(X+Y)={\rm var}(X)+{\rm var}(X)$ when $X,Y$ are independent, Jun 29, 2012 at 3:00