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I have an integral-normalised probability distribution $P_x$ that looks like this:

enter image description here

The peak of this distribution occurs at $x\approx5.2$. I am trying to find a way to measure the uncertainty on this peak value.

Clearly the distribution is not Gaussian, so surely I cannot simply integrate under this curve to find where the area becomes $34%$ on each side of the peak. Furthermore, the area under the curve on the left-hand side of the peak is less than $34%$ anyway.

Any help is much appreciated.

Thanks

ajrlewis
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  • What does $34$ have to do with this? Is the total area $68$? The function isn't flattened off by $x=10$. – J.G. Jul 13 '18 at 11:52
  • @J.G. Thanks for getting back to me. The total area of this curve is $100%$, even though as you say it does not flatten off at $x = 10$. The $34%$ corresponds to the area expected from the peak to the $1\text{-}\sigma$ value assuming that this was a Gaussian distribution. – ajrlewis Jul 13 '18 at 11:57

1 Answers1

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I think you've got this backwards. I'll assume you've got the numerical integration skills to compute the mean and variance; your problem, I think, is you're not using them, and want to instead infer some kind of "effective $\sigma$" from probabilities.

Why does anyone care about $1\sigma$ with a Gaussian distribution? Well, the fact we know in that case the probability of being within such a confidence interval is convenient, but we mainly like it because, well, why wouldn't you like to count in $\sigma$s?

Why do we care about a $68%$-wide CI? Not because anyone cares about that number, but because for a Gaussian distribution that's what $\pm 1\sigma$ is equivalent to.

For an arbitrary distribution, as long as you can do integration (perhaps numerically) you can have any CI you want. But you have to decide whether you want to define it by width or probability. If you want to specify the mean and SD, or equivalently $\mu\pm\sigma$. go ahead. If you want to find the unique $z>0$ for which $P(|X-\mu|\le z\sigma)=0.68$, do that instead. I'm pretty sure you don't actually care about that value! (Fortunately, the next famous result, that $P(|X-\mu|\le 2\sigma)\approx 0.95$ for a Gaussian, has a nicer-looking RHS.)

Since the distribution isn't symmetric, maybe you'd rather quote a CI around the median. I'll leave that to you.

J.G.
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  • Thanks for your answer. So, if I want to keep the peak value, which is neither the mean nor median in this example by the way, I can just find the CI of 68% from this peak value? However, on the LHS, there is not enough area to determine the where the 34% value occurs. Therefore, is the lower 68% CI not defined in this example? Thanks in advance – ajrlewis Jul 13 '18 at 12:18
  • or can I just quote the peak value and the standard deviation of this distribution, even though it isn't Gaussian? – ajrlewis Jul 13 '18 at 12:19
  • @AlexJ.R.Lewis I recommend doing what you can with the information at your disposal, whether that means using the median, mean or mode as your centre, and using whatever width or probability you're interested in (unless your data's insufficient for it). Just make what you're doing explicit and you should be fine. In fact, if you need to communicate to someone else, the chart is worth showing to them so they can see how difficult doing this was! – J.G. Jul 13 '18 at 12:29
  • Quite -- thanks for your help on this issue. – ajrlewis Jul 13 '18 at 12:31