Timeline for How to fit one distribution to another?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Sep 16, 2012 at 15:40 | comment | added | whuber | No, that's incorrect. My claim is that the global extrema of $|F_1(x)-F_2(x)|$ are to be found among the critical points of $(F_1(x)-F_2(x))$ provided $F_1-F_2$ is differentiable. This should be obvious when you consider the effects of squaring both sides. | |
| Sep 15, 2012 at 7:05 | comment | added | Sjoerd C. de Vries | @whuber So it is your claim that $\frac{d}{d\lambda}sup_x|cdf1(x)-cdf2(x,\lambda)|=sup_x|\frac{d}{d\lambda}(cdf1(x)-cdf2(x,\lambda))|$ | |
| Sep 14, 2012 at 22:35 | comment | added | whuber | Good points. (1) The lack of dependence of the data-based CDF on $\lambda$ is one indication that this approach is not a good (or even valid) way to assess the data distribution. (See my comment to the question itself.) (2) Avoid the nondifferentiability problem by differentiating the difference of the CDFs (which is differentiable) and looking for critical points: you will find the upper and lower Kolmogorov distances among them, then you can choose the largest in absolute value. | |
| Sep 14, 2012 at 22:32 | comment | added | Sjoerd C. de Vries | @whuber Not sure that's correct. First, shouldn't you differentiate wrt $\lambda$? Well, one of the CDFs isn't even dependent on $\lambda$. And second, differentiating the sup of absolute differences doesn't equal the sup of the absolute differences of the derivatives, does it? | |
| Sep 14, 2012 at 15:45 | comment | added | whuber | Working with the Kolmogorov distance is not a problem, because we're looking at the differences between the CDFs, which--as integrals of the PDFs--are automatically differentiable (at least for absolutely continuous variables). Not only that, you automatically have the derivative available: it's the PDF! So just find zeros of the differences of the two PDFs. | |
| Sep 14, 2012 at 8:33 | comment | added | Sjoerd C. de Vries | @whuber In fact, the K distance may be troublesome in this case, because of the strongly non-linear $sup$ function involved. Since finding the minimum difference between the distributions involves differentiating it wrt $\lambda$ that could be a problem. Do you know how to solve that? | |
| Sep 14, 2012 at 5:51 | comment | added | Sjoerd C. de Vries | @whuber I don't claim any expertise in this area, so I must assume you're right here. Not that I think it would help, as the K distance is probably as difficult to calculate in this case as the $L^2$ norm. So, the numerical route may be needed anyway. | |
| Sep 13, 2012 at 22:17 | comment | added | whuber | The $L^2$ norm for the PDFs is not usually an appropriate or useful measure of differences between distributions. Look instead at the $L^1$ norm for the CDFs (the Kolmogorov distance) or the Kullback-Leibler divergence, etc. These issues are extensively discussed in questions at stats.stackexchange.com. | |
| Sep 13, 2012 at 21:19 | history | answered | Sjoerd C. de Vries | CC BY-SA 3.0 |