Timeline for Moment Generating Function of a nonlinear transformation of an exponential random variable
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 16, 2014 at 21:41 | vote | accept | Dennis | ||
| Sep 16, 2014 at 21:24 | comment | added | Dennis | The Legendre Transform of $\log MGF_V$ is the instantaneous cost function in an optimal control problem. So I need to differentiate it and invert it at a few steps, as well as integrate it over a path $v(t)$ to evaluate the cost of the path. Doing this numerically in Matlab has been a huge headache because depending on the parameters ($\delta$,$\lambda$) numerically solving equations using them doesn't always give (what I believe to be) correct answers. I'm happy to be educated on alternate ways to express it! | |
| Sep 16, 2014 at 20:40 | comment | added | whuber♦ | (1) For the record, the correct formula for $\text{ExpIntegralE}(n,t)$ is $\int_1^\infty \exp(-t x) /x^n dx$. Differentiating it with respect to $t$ is easy, because it just produces another exponential integral. Differentiation with respect to $n$ is very messy. (2) It's not evident where the "instability issues" are: could you be more specific about that? (3) The Gamma function representation looks particularly nice and is straightforward to differentiate with respect to $t$. | |
| Sep 16, 2014 at 20:34 | answer | added | Alecos Papadopoulos | timeline score: 3 | |
| Sep 16, 2014 at 19:10 | history | edited | Dennis | CC BY-SA 3.0 |
Specifying range of integration
|
| Sep 16, 2014 at 18:58 | comment | added | Aniko | You are right, I missed the fact that the limits become 0 to 1 after the variable change. | |
| Sep 16, 2014 at 18:58 | comment | added | Dennis | I don't know why Mathematica comes up with the forms it does. If I simplify the density, letting $A=\frac{\lambda}{\log \delta}$, and reducing things like $Exp[-A Log[v]]= v^{-A}$, it will give the expression in the later part of my note. | |
| Sep 16, 2014 at 18:56 | comment | added | Dennis | I didn't specify the limits of integration, but for $\tau$ it is 0 to inf, and after the change of variable for $v$ it is from 0 to 1 (with a sign change), since the density is 0 elsewhere. When I calculate the MGF at particular values, it does have the properties I expect, so I think it is correct. The problems come when I want to invert it, take derivatives, and do all this other stuff later in my problem, and it becomes a nightmare. | |
| Sep 16, 2014 at 18:51 | comment | added | Aniko | Are you sure that you are calculating the definite integral (0 to infty), and not just the indefinite integral? I don't get the incomplete Gamma function in there. | |
| Sep 16, 2014 at 17:45 | review | First posts | |||
| Sep 16, 2014 at 17:49 | |||||
| Sep 16, 2014 at 17:40 | history | asked | Dennis | CC BY-SA 3.0 |