Timeline for What conditions should be satisfied for a rational function to be a moment generating function?
Current License: CC BY-SA 4.0
15 events
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| Jan 8 at 9:18 | vote | accept | Paul R | ||
| Jan 5 at 18:28 | comment | added | Paul R | Thank you very much! | |
| Jan 5 at 17:19 | comment | added | Iosif Pinelis | @PaulR : Your approach -- to first fit the data to some rational function that is not an m.g.f. -- was not a good idea at all. The much better idea is to fit the data right away to an m.g.f. that is rational over the range of the data. This also answers your stated question: ""find a rational function that can be a moment generating function". Also, I have now simplified the answer, quite a bit, with a simple and explicit expression for the p.d.f., and the fit is now almost perfect. | |
| Jan 5 at 17:06 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jan 5 at 16:40 | comment | added | Paul R | I just managed to approximate my tabulated data by some rational function. And then arised a question is this rational function an MGF. So, you answered that rational function can be an MGF. But what I need is to approximate my data with rational function. It is well done with any rational function, but I want to approximate it with rational MGF. | |
| Jan 5 at 16:28 | comment | added | Iosif Pinelis | @PaulR : (i) Your question was to "find a rational function that can be a moment generating function". This is done in formulas (1) and (2). (ii) Never before you had asked "Can we really approximate a rational function by a sum of non-negative rational functions?". And indeed that is not a problem, because any m.g.f. is $>0$; so, any rational m.g.f. is already a nonnegative rational function. | |
| Jan 5 at 16:10 | comment | added | Paul R | UPDATE: I've found an article about mixture distribution. The question is the following: if MGF is defined for $t\lt\frac{1}{b}$ then $(1-bt)^{-a}$ is positive. Can we really approximate a rational function by a sum of non-negative rational functions? | |
| Jan 5 at 15:53 | comment | added | Paul R | Possibly, I don't know what the word "mixture" here means. From my point of view it is just a linear combination of MGFs. | |
| Jan 5 at 15:15 | comment | added | Iosif Pinelis | @PaulR : The p.d.f. corresponding to the mixture (not the sum) of m.g.f.'s is the mixture (not the sum) of p.d.f.'s corresponding to the mixed m.g.f.'s. This follows from the formula $M(t)=\int_{-\infty}^\infty e^{tx}f(x)\,dx$, where $M$ is the m.g.f. and $f$ is the corresponding p.d.f. You seem to be confusing this with the completely unrelated fact that the m.g.f. of the sum of independent r.v.'s (not of the sum of p.d.f.'s!) is the product of the m.g.f.'s of the r.v.'s. I have now added these details to the answer. | |
| Jan 5 at 14:49 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jan 5 at 10:59 | comment | added | Paul R | Thank you very much! We assume that empirical MGF is sum of the MGFs of the Gamma distributions. But how do I recreate the PDF? Sum of PDFs is the product of MGF's, not sum. | |
| Jan 5 at 4:21 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jan 5 at 4:01 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jan 4 at 23:10 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jan 4 at 22:53 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |