I am familiar with the following condition for statistical independence where $x$ and $y$ are two different (random) variables: \begin{equation} E(xy)=E(x)\times E(y) \end{equation}

Now, I came across a paper (Appendix B, p.1311) where it mentions that if $h$ and $^{\partial h}/_{\partial z}$ are independent then following holds: \begin{equation} E\left(h \cdot \frac{\partial h}{\partial z}\right)=E(h)\times E\left(\frac{\partial h}{\partial z}\right) \end{equation}

I would like to know: What does it mean if $h$ is a dependent, continuous variable--such as smoking expenditure (in dollars)--and $z$ is a continuous independent variable--say income in dollars--in regression model. Note that $^{\partial h}/_{\partial z}$ is the partial derivative of $h$ with respect to $z$ which is equal to $dh/h$ (percentage change in $h$)

  • $\begingroup$ Maybe you should provide the precise definition of $h$. The expectation of a derivative is a quite unusual thing. $\endgroup$ – Stéphane Laurent Sep 21 '13 at 15:40
  • $\begingroup$ @ Stephane : I have updated that; hope it is now more clear $\endgroup$ – user227710 Sep 21 '13 at 15:59
  • $\begingroup$ @StéphaneLaurent, is it so unusual? ;) user227710, it is unclear what the notation $E(h,\partial h/\partial z)$ means. In particular, the comma is confusing. Do you mean $E( h \cdot \partial h/\partial z)$? Also, if this is an ordinary linear regression model, the mean structure is fixed and the the error term is the only source of randomness, so if $z$ is independent of the error, then $\partial h/\partial z$ would be a fixed constant and can be pulled out of $E( h \cdot \partial h/\partial z)$. Is that what this question is about? $\endgroup$ – Macro Sep 21 '13 at 17:05
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    $\begingroup$ $E(xy)=E(x)E(y)$ is zero correlation, not independence. $\endgroup$ – Glen_b Sep 22 '13 at 4:11
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    $\begingroup$ @Glen_b 's point is correct and seriously reorients the scope of this question. It's worth noting that only under a very strict assumption of jointly normally distributed variables does 0 correlation imply independence. Otherwise, independence is defined using the probability density. See Casella Berger (2001) $\endgroup$ – AdamO Sep 22 '13 at 4:56

The marginal change of a variable w.r.t. to another variable cannot be equal to the corresponding percentage change of the former. The (point) percentage change equals the total differential of the logarithm of a variable, $dln(y) = dy/y$

The appendix you refer to writes (using your letters) $E[h\cdot (dh/h)]$, where $(dh/h)$ is the relative (marginal percentage) change in $h$ due to a change in some explanatory variable $z$. $d$ symbolizes total differential.

Then the paper says that if $h,\; (dh/h)$ are independent, then as usual the expected value can be broken. For these two components to be independent, it must be the case that $h$ does not really appear in $(dh/h)$. This requires that $dh$ is a linear function of $h$, something like $$dh = ahdz \Rightarrow dh/h = adz \qquad (\Rightarrow \frac{dh}{dz} = ah)$$

Related to the economic meaning of this, it is that the total marginal effect $z$ has on $h$ depends on the level of $h$. Using your example, if $h$ is smoking expenditure and $z$ is income, then:

-if $a$ is positive, it would mean that the higher smoking expenditure is, the more it increases with income.

-if $a$ is negative, it would mean that the higher smoking expenditure is, the less it increases with income.


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