I'm a little puzzled by the whole random variable thing.

I've got two random variables, $\mathcal{X}$ and $\mathcal{N}$, both with gaussian distribution with mean = 0 and $\sigma_{\mathcal{X}}^2$ and $\sigma_{\mathcal{N}}^2$ respectively.

The equation for correlation coefficient is

$$ \rho_{XY} = E\left[\frac{X-E[X]}{\sigma_{\mathcal{X}}}\frac{Y-E[Y]}{\sigma_{\mathcal{N}}}\right] $$

I know how to find $E$ but what value do $\mathcal{X}$ and $\mathcal{N}$ actually have that I plug into the equation?

  • $\begingroup$ The thing inside $[\cdot]$ is another random variable, and $E$ denotes its expectation value. You're not supposed to plug in any single values of $\mathcal{X}$ and $\mathcal{N}$. $\endgroup$ – JiK Feb 25 '14 at 22:16
  • $\begingroup$ The trouble is that that $E$ is an expectation over $X$ and $Y$ jointly. You need, e.g., the covariance as well to be able to calculate the correlation. $\endgroup$ – TooTone Feb 25 '14 at 22:40

When you calculate an expectation ($E$) of a random variable $Z$ you sum up for all values of $Z$ times the respective prbability of $Z$ taking that value. Accordingly, you work here. So every possible value of $X,Y$ will apper (be plugged) in the equation, multiplied however with the repsective probability of occuring (see formula of expected value).

If you know that two random variables are independent then that implies that they are uncorrelated. (independent implies uncorrelated but not the other way round). So in that case $$\rho_{XY}=0$$ and you do not need to apply the definition nor do any calculations to prove it.

  • $\begingroup$ is the coefficient not simply zero? Since $E[\frac{XY}{\sigma_X\sigma_Y}] = 0$ $\endgroup$ – dingari Feb 26 '14 at 10:42
  • $\begingroup$ They are independent, I failed to note that in my initial post. What does that tell us? $\endgroup$ – dingari Feb 26 '14 at 11:04
  • $\begingroup$ Well now he assigned the same problem, but with the correction that we should calculate the correlation coefficient of X and Y. $\endgroup$ – dingari Mar 13 '14 at 20:44
  • $\begingroup$ X and N are independent but Y = X + N $\endgroup$ – dingari Mar 25 '14 at 17:43
  • $\begingroup$ @user2750354 Write correctly the question and you will get a reasonable answer.... You are changing all the time $\endgroup$ – Jimmy R. Mar 25 '14 at 17:51

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