Skip to main content
11 events
when toggle format what by license comment
Apr 21, 2023 at 13:21 history edited User1865345
edited tags
Apr 21, 2023 at 13:19 comment converted from answer User1865345 You can deal with a general result when you are considering convex combinations: if $\mathrm P_1, \mathrm P_2$ are probability generating functions and $\alpha\in[0, 1], $ then $\alpha\mathrm P_1 +(1-\alpha) \mathrm P_2$ is a probability generating function. It is not hard to see why: check whether the coefficients of $s^n$ in the convex combination (are non-negative, of course) sum to $1.$ Reference: Probability and Random Processes, Geoffrey Grimmett, David Stirzaker, Oxford University Press, 2020, sec. 5.1.
Jun 22, 2017 at 2:38 answer added Lucas Roberts timeline score: 2
Nov 10, 2015 at 21:42 history edited Glen_b CC BY-SA 3.0
added 166 characters in body
Nov 10, 2015 at 21:41 comment added Glen_b That independence assumption is critical; I'll edit this information into your question.
Nov 10, 2015 at 17:37 comment added jje Sorry for being unclear. I was trying to ask about $G_{X+Y}$, so the PGF of the sum of $X$ and $Y$. The variables can be assumed to be independent.
Nov 10, 2015 at 14:28 comment added Glen_b The sum of two functions is well defined (as is the difference); if you say "the sum of $G_X$ and $G_Y$" you're simply saying "$G_X+G_Y$" in words. What is it you actually mean to ask? Are you asking for the pgf of $X+Y$ (ie the pgf of the sum, rather than the sum of the pgfs?)
Nov 10, 2015 at 9:34 answer added jtobin timeline score: 4
Nov 10, 2015 at 9:30 answer added Olivier Hubert timeline score: 1
Nov 10, 2015 at 9:08 review First posts
Nov 10, 2015 at 9:56
Nov 10, 2015 at 9:08 history asked jje CC BY-SA 3.0