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Let's say I have a histogram with counts $y$ high enough to use $\Delta y = \sqrt{y}$ as error bar. If I transform this histogram in (natural) logarithmic scale, I receive as error bars:

$$ z = \log(y) \Rightarrow \Delta z = \frac{1}{y} \Delta y = \frac{1}{y} \sqrt{y} = \frac{1}{\sqrt{y}} $$

Now what I don't understand is: In the original scale my error $\Delta y$ goes $\bf{up}$ if my bin-count $y$ goes up. In logarithmic scale, my error $\Delta z$ goes $\bf{\text{down}}$ if my bin-count y goes up. How is this possible?

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    $\begingroup$ error bar = $\Delta y$ = $\sqrt{y}$ ??? $\endgroup$
    – ocram
    Oct 7, 2016 at 11:33
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    $\begingroup$ @ocram common in high-energy physics when poisson-distribution applies: science20.com/quantum_diaries_survivor/… $\endgroup$
    – Christian
    Oct 7, 2016 at 11:45
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    $\begingroup$ OK, I see. $Y \sim$ Poisson. Thus $E(Y) = Var(Y)$. And $Var(log(Y)) \approx \frac{1}{E(Y)}$ (delta method). $\endgroup$
    – ocram
    Oct 7, 2016 at 12:02
  • $\begingroup$ A key point to figure this out intuitively, I think, is that, using the logarithmic transformation, large counts are shrunk, and the extent of shrinkage is greater as the counts get larger $\endgroup$
    – ocram
    Oct 7, 2016 at 12:39

2 Answers 2

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The error bar appears to be shorter because the same range requires less space higher up on the graph (where the divisions are closer together). Divisions on the logarithm scale come together as $\frac{1}{y}$ (the gradient of the logarithm) which will decrease the apparent length of the error bars (which in your case increase with $\sqrt{y}$)

So - the error doesn't get smaller, but it is represented by a shorter bar because everything is closer together high up in the log plot. The distance between 1 and 2 is the same as the distance between 100 and 200. That's just how log plots work.

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If you start with $x$ counts and you want to display $ x \pm \sqrt{x}$ then your error bar is of length $2\sqrt{x}$.

If instead your count is $kx$ counts and you want to display $ kx \pm \sqrt{kx}$ then your error bar is of length $2\sqrt{kx}$.

So your error bar on the larger count is longer than the original error bar by a factor of $\sqrt{k}$, even though it is a smaller proportion of the new larger count also by a factor of $\sqrt{k}$.

And it is that smaller proportion which is being shown on your log scale. Your original error bar would have length (proportional to) $\log_e( x+\sqrt{x})-\log_e( x-\sqrt{x}) \approx \frac{2}{\sqrt{x}}$ on the log scale while for the larger count the length is (proportional to) about $\frac{2}{\sqrt{kx}}$, now actually shorter by a factor of about $\sqrt{k}$

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