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Fitting a pdf against Weibull pdf

Moment matching is a standard tool. Whether it works well needs to be determined. The mean $mu_1$ and the next two central moments $mu_2$ and $mu_3$ of this mixture-of-exponentials PDF are$$eqalign mu_1 &= fracfracsbfraca-basa-bs mu_2 &= fracb^2s (2 as)b^2 (as)^2 mu_3 &= frac2 left(1frac-a^3b^3(as)^3

ight)b^3. $$Equate these with the corresponding moments of the Weibull distribution (which equal $442626$, $1.56725 times 10^11$, and $1.04716 times 10^17$, respectively) and solve. The general solution is a little messy but easy to compute:$$eqalign s &= frac4 mu_1^3-6 mu_1 mu_2mu_3 pm sqrt-2 left(mu_1^6-3 mu_1^4 mu_29 mu_1^2 mu_2^2-9 mu_2^3

ight)4 mu_1 left(2 mu_1^2-3 mu_2

ight) mu_3mu_3^2mu_1^43 mu_2^2-2 mu_1 mu_3 a &= -frac2 left(mu _1^3-3 mu _1 mu _2mu _3

ight)mu _1^43 mu _2^2-2 mu _1 mu _3 b &= pm frac2 mu _1^3- mu _3-sqrtleft(4 mu _1^3-2 mu _3

ight)^2-24 left(mu _1^2-mu _2

ight) left(mu _1^43 mu _2^2-2 mu _1 mu _3

ight)mu _1^43 mu _2^2-2 mu _1 mu _3$$(Notice that all the denominators are the same.)The two solutions differ by interchanging $b$ with $as$, as would be expected from the symmetry in the pdf: they are two distinct parameterizations of the same pdf.The solutions for this particular Weibull distribution turn out to be $a = 1.72179 times 10^-6$ and either (i) $s = 2.49314 times 10^-6$, $b = 2.88012 times 10^-6$ or, equivalently, (ii) $s = 1.15833times 10^-6$, $b = 4.21493times 10^-6$. In the figure, the blue line shows the PDF of the Weibull distribution and the red line shows the mixture-of-exponentials fit:The distributions look a little different for small values. Here is a magnified version near 0:The mixture of exponentials just cannot reproduce the behavior of a Weibull distribution in this range. If the fit will be used to assess probabilities in the right tail, though, we might be ok.Another natural way to compare two distributions is by the vertical deviations between their CDFs:At this scale, the two look coincident: this is good. To appreciate the differences better, look at the differences between the two CDFs (fitted minus actual):They never differ by more than 0.006 or less than -0.003. Therefore, when using the CDFs to compute probabilities of any interval, the results will never be wrong by more than 0.006 - (-0.003) = 0.009, less than 1%. The differences in the right tails apparently go to zero, indicating the fit ought to be very good for right tail probabilities.