Timeline for Moment Generating Function of a nonlinear transformation of an exponential random variable
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Sep 16, 2014 at 22:13 | comment | added | Dennis | Sure, I'll do that. | |
| Sep 16, 2014 at 22:04 | comment | added | Alecos Papadopoulos | Then perhaps, if you update your question in math.SE using the results from my answer, and the "equation of the type..." you typed in your comment above, perhaps you will get some useful feedback there, adjusting perhaps also the tags of the question. | |
| Sep 16, 2014 at 22:00 | comment | added | Dennis | Well, certainly. As I said, I asked originally in math.stackexchange.com | |
| Sep 16, 2014 at 21:58 | comment | added | Alecos Papadopoulos | It appears that your interest in the MGF is not so much in its capacity as a generator of distribution moments (which happens when $t=0$), since you want to solve for $t$. This apparently makes your problem much more mathematical than statistical, don't you think? | |
| Sep 16, 2014 at 21:51 | comment | added | Dennis | Thanks very much! I think this answer(and whuber's comment) will help a lot. Other than typical equation solving approaches, any advice for solving an equation of the type: $\frac{MGF_V^{\prime}(t)}{MGF_V(t)} - s =0$ for $t$? This was the sort of exercise that gives me trouble, numerically. What I had been doing was using implementations of the $\Gamma$ function and then using either built in or home-cooked equation solving methods, having a lot of trouble as parameters and $s$ varied. | |
| Sep 16, 2014 at 21:41 | vote | accept | Dennis | ||
| Sep 16, 2014 at 21:27 | comment | added | Alecos Papadopoulos | @whuber Thanks. Indeed, there lies the general case. | |
| Sep 16, 2014 at 21:25 | history | edited | Alecos Papadopoulos | CC BY-SA 3.0 |
added 98 characters in body
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| Sep 16, 2014 at 20:54 | comment | added | whuber♦ | +1 Very nice! You have also incidentally provided another answer to stats.stackexchange.com/questions/114481 for the case $\lambda/\log(\delta)=-2$. Note that the MGF is a confluent hypergeometric function, which makes its computation susceptible to a huge array of techniques. | |
| Sep 16, 2014 at 20:34 | history | answered | Alecos Papadopoulos | CC BY-SA 3.0 |