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Let $X$ be a random variable, and $f$ a measurable function. Is there any particular relationship between the expression of $f$ and $corr(f(X),X)$?


The background of asking the value of $corr(f(X),X)$ is as following.

From the book on elementary statistics, I learned the conditional expection $E[Y|X]$ is the best approximation to the random variable $Y$, under the criterion of minimizing the $E[Y-L(X)]^2$.

Denote by $M(X)$ the $E[Y|X]$, and assume $(X,Y)$ subject to the multivariable normal distribution with parameters(mean, variance and correlation) $u_1,u_2,\sigma_X^2,\sigma_Y^2,\rho$.

In this case, ` $M(X)=u_2+\frac{\rho \sigma_2}{\sigma_1}(X - u_1)$. And computation shows $$ corr(M(X),X) =\frac{ \sigma_X}{ \sigma_M} M'(0) $$. So this seems a nice relation.


So I just want to know, is it possible to extend the above result to general $(X,f(X))$?

Or, if we restrict $f$ to a special class of functions, is there any nice result on $corr(X,f(X)$? I tried write $f(X)$ as a power series, but came up with nothing interesting.

If there are some results on it, where can I find the references?

Thank you.

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up vote 1 down vote accepted

At least there is a simple result for the case $f(x)=e^{tx}$. Under suitable conditions, we have $$ {\rm corr}(X,e^{tX} ) = \frac{{{\rm E}[Xe^{tX} ] - \mu _X {\rm E}[e^{tX} ]}}{{\sigma _X \sqrt {{\rm E}[e^{2tX} ] - {\rm E}^2 [e^{tX} ]} }} = \frac{{m'_X (t) - \mu _X m_X (t)}}{{\sigma _X \sqrt {m_X (2t) - m_X^2 (t)} }}, $$ where $m_X (\cdot)$, $\mu_X$, and $\sigma^2_X$ denote the moment-generating function, the expectation, and the variance of $X$, respectively.

For example, for $X$ exponential with mean $1/\lambda$ (hence $m_X (\tau) = \frac{\lambda }{{\lambda - \tau}}$, $\tau \lt \lambda$, $\sigma_X = 1/\lambda$), this yields $$ {\rm corr}(X,e^{tX} ) = \frac{{\lambda t}}{{(\lambda - t)^2 \sqrt {\frac{\lambda }{{\lambda - 2t}} - (\frac{\lambda }{{\lambda - t}})^2 } }} , \;\; t \in ( - \infty ,\lambda /2)\backslash \lbrace 0 \rbrace . $$ (Using ${\rm E}[X^n] = n!/\lambda^n$, we can also find that ${\rm corr}(X,X^n) = \frac{n}{{\sqrt {{2n \choose n} - 1} }}$.) Note that $\lim _{t \to 0 \pm } {\rm corr}(X,e^{tX} ) = \pm 1$ and $\lim _{t \uparrow \lambda /2} {\rm corr}(X,e^{tX} ) = 0$. (For the former, consider $e^{tX} \approx 1 + tX$.)

The general formula for ${\rm corr}(X,e^{tX} )$ may help you reach some conclusion. In particular, note that the functions $f(x)=e^{tx}$ and $m_X (t)$ are often closely related, see here, for example.

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Very interesting, I didn't think of computing $X$ v.s. its own generating function. I love the formula $ lim_{t\rightarrow 0^+} {\mathrm{corr}}(X,e^{tX})=1 $. I thought it's nice and should be correct for all distributions. Because when $t$ goes to 0, we have the following approximation: $$ \frac{e^{tX}-1}{\sqrt{Ee^{2tX}-(Ee^{tX})^2}} = \frac{ X+O(t)}{\sigma_X \sqrt{1+O(t)} } $$. It confirms the fact that correlation is a linear dependence measurement between two random variables. – J.Xie Feb 24 '11 at 14:45
Thank you. For $\lim _{t \to 0 \pm } {\rm corr}(X,e^{tX} ) = \pm 1$, I considered the fact that ${\rm corr}(X,1 + tX) = \frac{t}{{|t|}}$. – Shai Covo Feb 24 '11 at 16:28

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