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Let’s say I have got 15 CMIP5 models’ outputs of a given variable e.g. temperature, for RCP6.5 (2020-2100) run.

I want to plot how the temperature will change in the future 2020-2100 (scatter plot, x = time and y = temperature).

Instead of showing all the 15 models outputs can I just simply show the CMIP5 multi-model ensemble average?

i.e. instead of 15 lines into one plot I show only one line (the average).

Is this correct? Will a reviewer reject this approach?

Thanks

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    $\begingroup$ I don't have any familiarity at all with being reviewed, nor with the model. But would think perhaps you could show "error bars" to indicate the variance over the members. The question seems to be why you wish to eliminate the multiple lines... if it's to mask out outliers that show different outcomes, seems poor science... but if it's just because the 15 lines will make readability difficult, certainly fair enough... and perhaps a suggestion like given is a good idea to maintain the detail without the reading challenge :-) $\endgroup$ – JeopardyTempest Feb 24 '18 at 15:01
  • $\begingroup$ The reason why I want to show the ensemble average is indeed because I want to get rid of such many lines that create confusion. Thanks for the suggestion. $\endgroup$ – aaaaa Feb 24 '18 at 15:12
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    $\begingroup$ Shaded area showing the spread should be shown with the avg. You could also include all members as faint grey lines. $\endgroup$ – farrenthorpe Feb 24 '18 at 15:17
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    $\begingroup$ Your concerns are legitimate. A way to mitigate them is to plot anomalies from an average over a reference time period. This focuses the plot on temperature change, which you said is what you want to show. $\endgroup$ – K. Lindsay Feb 24 '18 at 15:31
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    $\begingroup$ I guess it depends on your objective. In forecasting, the average is very useful. $\endgroup$ – farrenthorpe Feb 24 '18 at 15:31
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That is valid. You can show the average (or median if there is a reason to get rid of outliers), but it would be VERY misleading and arguably wrong if you don't show also the uncertainty associated with that mean solution. However, the whole point of getting the mean of many values is to reduce the uncertainty. Therefore, just plotting a shaded area covering the range over which the original lines were distributed, would be an overestimation of the uncertainty (here I'm assuming that each original line have no uncertainty, which is wrong, but if you have that uncertainty data it can also be included). A standard way to estimate the uncertainty of the average of many lines (a.k.a Time series) is calculating bootstrap confidence intervals.

Here is an example from a paper I recently submitted: enter image description here

The mean of all those black lines is the blue line. To expose the spread of the data I decided to show all the original lines. But, if I were to show this to a general audience, I would get rid of the thin black lines and show the mean (blue) and the confidence intervals (pink shades). Those shades mean that with a 90% confidence the mean should fall within the dark pink shading, and with a 99% confidence within the light pink shading. Giving a sense of how much you can trust the blue line.

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  • $\begingroup$ Thank you. Is there any R package that performs the calculation of bootstrap confidence intervals? Also, why not just plotting the standard error of the mean as shaded area? $\endgroup$ – aaaaa Feb 25 '18 at 17:14
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    $\begingroup$ @aaaaa I'm sure there is, because is something fairly common. But I'm not an R user, so I don't know any details about that. I've done it in Matlab if that's of any help. $\endgroup$ – Camilo Rada Feb 25 '18 at 17:19
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    $\begingroup$ @aaaaa Standard error is also an option, but the meaning of it is much more difficult to grasp for non scientist (and even for scientists). Instead, saying that you have a 99% confidence that the line lies withing some boundaries is pretty straightforward to understand. $\endgroup$ – Camilo Rada Feb 25 '18 at 17:22
  • $\begingroup$ Ok. Suggestion accepted. I will have a play :) $\endgroup$ – aaaaa Feb 25 '18 at 17:28
  • $\begingroup$ @aaaaa Glad to hear that. If this answer is satisfactory please don't forget to accept it. $\endgroup$ – Camilo Rada Feb 25 '18 at 17:31

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