Brian Jackson edited Ultimately_though_we_re_interested__.tex  almost 9 years ago

Commit id: 01b90ef219bf21a7519d3184a2b68a76fa9ff43f

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\end{equation}  where $\left( \frac{d}{dP_{\rm obs}} \right)_{P_{\rm obs} = P_{\rm act}}$ means calculate the $P_{\rm obs}$-derivative of the expression in square brackets and then replace $P_{\rm obs}$ with $P_{\rm act}$. It is important to re-iterate that Equation \ref{eqn:n-Pact_from_n-Pobs} assumes $\Gamma_{\rm act}$ independent of $P_{\rm act}$, a limitation upon which we will improve below.   It's clear that applying Equation \ref{eqn:n-Pact_from_n-Pobs} to the $n(P_{\rm obs})$ from Equation \ref{eqn:n-obs_from_uniform_n-Pact} recovers the original uniform distribution. On the surface, though, Equation \ref{eqn:n-Pact_from_n-Pobs} suggests the strange result that for the power-law distribution $n(P_{\rm obs}) \sim P_{\rm obs}^{-2}$, similar to that reported in \citet{Jackson_2015}, the underlying distribution $n(P_{\rm obs}) = 0$. However, the approach here assumes that $P_{\rm obs}^{-2}$ only span a finite range, which is violated by the simple $P_{\rm obs}^{-2}$ power-law. Moreover, as we show below, including consideration of the a  distribution of $\Gamma_{\rm act}$-values significantly modifies the relationship between $n(P_{\rm act})$ and $n(P_{\rm obs})$.