Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have two random variables $X$ and $Y$ with mean and standard deviation $(\mu_1,\sigma_1)$ and $(\mu_2,\sigma_2)$ respectively. I know that for perfect correlation the relationship is given by a linear regression. I also know that positive perfect correlation establishes that $\mu_2$ will be linear function of $\mu_1$. But is there a relationship between variances ? More specifically given $\mu_1,\sigma_1$ and $\mu_2$, can I evaluate what is the value of $\sigma_2$ ? You can assume normal distributions for $X$ and $Y$ if that helps.

share|cite|improve this question

If two r.v.s $X,Y$ are perfectly correlated, $Y = mX+c$ for some constants $m,c$. So, $\mathbf{var}(Y) = m^2\mathbf{var}(X)$. Also $\mu_Y = m\mu_X+c$

share|cite|improve this answer
Can you compute m in terms of mu1, sigma1, and mu2 ? – Amit Dec 6 '12 at 1:28
I don't think so. m is usually obtained by regression. – dexter04 Dec 6 '12 at 1:31
Given mu1,mu2 and sigma1, there are infinitely many possible values of m, depending on sigma2. – dexter04 Dec 6 '12 at 1:33
Isn't the m governed by the correlation coefficient or is it dependent on sigma ? I thought that for perfect positive correlation m > 0. – Amit Dec 6 '12 at 1:41
@Amit certainly $m>0$ if there is perfect positive correlation, but it could be any number greater than zero. There's no way to calculate it with only the information given. – Jonathan Christensen Dec 6 '12 at 2:07

If $X$ and $Y$ are perfectly positively correlated random variables, then all the probability mass lies on a line of slope $\frac{\sigma_Y}{\sigma_X}$ passing through the point $(\mu_X, \mu_Y)$. So, if you know the values of $\mu_X, \sigma_X$ anfd $\mu_Y$, you know one point on the line but cannot determine its slope (and hence the value of $\sigma_Y$) from just this information.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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