Suppose I have a uniform random variable $X$ taking values $\{1,...,n\}$, and two functions $v(X)$ and $w(X)$.

I know that $v(X)$ and $w(X)$ are jointly distributed with correlation $\rho$. (And can assume any distribution structure including joint normal if it makes it easier).

First, I observe all values of $v(X)$ for $x=\{1,...,n\}$ and estimate $E[v(X)]$ as the average of the observations.

My goal is to estimate $E[w(X)]$ without sampling $w(X)$ by using the values of $v(X)$, the mean $E[v(X)]$ and the correlation coefficient.

Any hints/directions will be appreciated, including a Bayesian approach.

  • $\begingroup$ Welcome to the site. You don't need to sign your post, as the site does it for you. Also, I spelled out r.v.; many readers have English as a 2nd (or 3rd or 4th) language and abbreviations can be hard to decipher in a foreign tongue. $\endgroup$
    – Peter Flom
    Nov 15, 2012 at 0:41
  • $\begingroup$ Thanks Peter! I assumed a stats site will have r.v.s as the main dish. :) $\endgroup$
    – Ron
    Nov 15, 2012 at 0:45
  • $\begingroup$ Well, probably MOST people, even those who don't speak English fluently, would know what you mean, but it doesn't hurt to make things as clear as possible (statistics is difficult enough even when it is clear!) $\endgroup$
    – Peter Flom
    Nov 15, 2012 at 0:53
  • $\begingroup$ Does $X$ have any specific distribution? $\endgroup$ Nov 15, 2012 at 0:57
  • $\begingroup$ $X$ is uniform if that helps. $\endgroup$
    – Ron
    Nov 15, 2012 at 1:00

1 Answer 1


No, you cannot say anything at all about $E[w(X)]$ from this.

Let $w_k(X)=w(X)+k$ for any constant $k$.

Then $v(X)$ and $w_k(X)$ have the same correlation $\rho$ as $v(X)$ and $w(X)$.

But this tells you nothing about $k$ or $E[w_k(X)]=E[w(X)]+k$, since $k$ can take any value.

  • $\begingroup$ That's a good point. I'm trying to think of a similar example where $k$ is not constant yet maintains the correlation. Nothing seems to come to mind - do you have one in mind? $\endgroup$
    – Ron
    Nov 15, 2012 at 4:34

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