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I am currently trying to understand how distributions emerge out of a simple calculation when combining data which have other distributional properties. What I mean with this is, as I am not a mathematician, is the following: Let's assume the following two equations

$\phi_{i,t}= \frac{S_{i,t}-C_{i,t}}{A_{i,t}}$ or $\phi_{i,t}=\frac{I_{i,t}}{A_{i,t}}$ where $S$ and $A$ are positive, $C$ is negative and $I$ can be positive and negative, too.

If I now plot the probability density function (PDF) over all $i$ in $t$ for the different variables and calculate $\phi$ a new "smoothed" distribution occurs. The plots look as follows: Plots

As I am not that familiar with the dynamics and properties lying behind that kind of process I would be very glad if someone could explain it to me or give me a hint where to look for the relevant literature.

Thank you in advance! Best, Alex

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  • $\begingroup$ I don't understand this: you define $I=S-C$ and say that $S$ is positive while $C$ is negative, and yet $I$ can be positive and negative??? $\endgroup$
    – Carlo Beenakker
    Commented Apr 12, 2019 at 14:25
  • $\begingroup$ Hi @CarloBeenakker thank you very much for your for comment. Yes, because $C$ can also be larger than $S$ and thus $I$ can be negative in some case. $\endgroup$
    – Alexander Hempfing
    Commented Apr 16, 2019 at 9:40
  • $\begingroup$ but surely, if $C$ is negative and $S$ is positive, as you write, then surely $I=S-C$ can only be positive???? (a positive number minus a negative number is always positive, isn't it?) $\endgroup$
    – Carlo Beenakker
    Commented Apr 16, 2019 at 11:27

1 Answer 1

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In order to find the distribution of $\phi$, you need to know the joint distribution of $S,A,C$. Then you can use the standard transformation technique, as described e.g. in Section 4.2; see also bibliography there, including Casella, G. and R.L. Berger. Statistical Inference. Pacific Grove, CA: Duxbury, 2002; see, in particular, page 185 there. You can also read Section 4.4.1, with several examples worked out there.

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  • $\begingroup$ Thank you @Iosif! I will have a look at all the information you have provided! $\endgroup$
    – Alexander Hempfing
    Commented Apr 16, 2019 at 9:41

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