2
$\begingroup$

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.

$\endgroup$

2 Answers 2

3
$\begingroup$

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 $\mu_Y = m\mu_X+c$

$\endgroup$
5
  • $\begingroup$ Can you compute m in terms of mu1, sigma1, and mu2 ? $\endgroup$
    – Amit
    Dec 6, 2012 at 1:28
  • $\begingroup$ I don't think so. m is usually obtained by regression. $\endgroup$
    – dexter04
    Dec 6, 2012 at 1:31
  • $\begingroup$ Given mu1,mu2 and sigma1, there are infinitely many possible values of m, depending on sigma2. $\endgroup$
    – dexter04
    Dec 6, 2012 at 1:33
  • $\begingroup$ Isn't the m governed by the correlation coefficient or is it dependent on sigma ? I thought that for perfect positive correlation m > 0. $\endgroup$
    – Amit
    Dec 6, 2012 at 1:41
  • $\begingroup$ @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. $\endgroup$ Dec 6, 2012 at 2:07
1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .