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