Timeline for Solving recursion / finding generating function of a probability mass function
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jan 22, 2017 at 20:26 | vote | accept | Matjaž Krnc | ||
| Mar 5, 2016 at 22:39 | vote | accept | Matjaž Krnc | ||
| Jan 22, 2017 at 20:26 | |||||
| Mar 5, 2016 at 14:21 | comment | added | SM2 | Yes, I'm pretty sure that $f_n(k)$ does not admit any nice representation. If you had one, you'd be able to put it into the sum which defines $F_n(x)$ and get a closed-form expression for the latter. As to the moments, you can get all of them in a subsequent manner. However, again you can't get a closed-form solution for all the moments at once. Esg mentioned one way to look at it, the only thing I can add is you can use more direct calculations. Just taking derivatives of $F_n(x)=qF_{n-1}(x)+pF^2_{n-1}(x)$ at x=1 will lead you to linear equations, that can be solved effectively. | |
| Mar 4, 2016 at 13:42 | comment | added | Matjaž Krnc | Thank you for a productive answer! I edited the question accordingly. As we are dealing with a generating function, we are usually never interested in $F_n(x)$ for specific points $x$, but rather for individual coefficients. That said, I would be also interested in a value of $\frac{\partial F_n}{\partial x}$ where $x=1$, which should be the same as the expected value of $k$ in the space with probability mass function $f_n$. Anyway; do you think that your answer implies that also $f_n(k)$ do not admit a closed-form representation? | |
| Mar 1, 2016 at 13:09 | history | edited | SM2 | CC BY-SA 3.0 |
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| Mar 1, 2016 at 12:04 | review | First posts | |||
| Mar 1, 2016 at 12:09 | |||||
| Mar 1, 2016 at 12:04 | history | answered | SM2 | CC BY-SA 3.0 |