enter image description here

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.

  • 2
    $\begingroup$ 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. $\endgroup$
    – user10525
    Apr 21, 2012 at 14:54
  • $\begingroup$ @Procrastinator I thought of what you wrote a bit and it makes sense. Thank you $\endgroup$
    – Andrew
    Apr 21, 2012 at 15:09
  • 1
    $\begingroup$ 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. $\endgroup$ Apr 23, 2012 at 15:46
  • $\begingroup$ @Andrew, this recent question answers this question in greater generality: stats.stackexchange.com/questions/31177/… $\endgroup$
    – Macro
    Jun 29, 2012 at 0:34

2 Answers 2


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.

  • 3
    $\begingroup$ +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, $\endgroup$
    – Macro
    Jun 29, 2012 at 3:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.