Timeline for How I can define a discrete probability distribution with parameters?

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Mar 23, 2017 at 21:15	answer	added	user44169		timeline score: 0
Mar 23, 2017 at 20:49	history	edited	user44169	CC BY-SA 3.0	added 1 character in body
Mar 23, 2017 at 20:02	history	edited	user44169	CC BY-SA 3.0	deleted 1 character in body
Mar 23, 2017 at 16:32	vote	accept	CommunityBot
Mar 23, 2017 at 16:30	history	edited	user44169	CC BY-SA 3.0	added 1 character in body
Mar 23, 2017 at 15:44	comment	added	JimB		I think that $K$ takes on the values 0 through $2 M$.
Mar 23, 2017 at 15:27	answer	added	JimB		timeline score: 4
Mar 23, 2017 at 13:33	comment	added	J. M.'s missing motivation♦		Right, I purposely did not give a complete definition; I'll leave that to somebody else. I only wanted to illustrate that `SeriesCoefficient[]` is useful for taking arbitrary-order derivatives of a generating function.
Mar 23, 2017 at 13:25	comment	added	user44169		@J.M. the series is finite in this case so `Coefficient`works, but the problem is that when I tried to use it inside `ProbabilityDistribution` it says that my distribution is zero. And I need all the coefficients, not only the negatives, as I showed in the piecewise function of the question.
Mar 23, 2017 at 13:11	comment	added	J. M.'s missing motivation♦		Actually, you'll want to use `SeriesCoefficient[]` for that: `f[k_Integer?Positive] := SeriesCoefficient[(c/(Dd x) + (Dd - t - c)/Dd + (t - d) x/Dd + d x^2/Dd)^m, {x, 0, k}]`. As for normalization: `Method -> "Normalize"` is useful for the lazy.
Mar 23, 2017 at 12:45	history	edited	user44169	CC BY-SA 3.0	added 157 characters in body
Mar 23, 2017 at 12:42	comment	added	user44169		the bracket mean "coefficient of". I tried to define this distribution with the built-in function `Coefficient`and using assumptions but it says that my distribution is zero... The useful thing here is that, if it works, I can have a symbolic expression (that depends on the parameters) for the mean.
Mar 23, 2017 at 12:12	comment	added	Szabolcs		Unfortunately I cannot write an answer using the example from your question because I do not fully understand it (I don't know the meaning of the square brackets in $[x]$ and I do not see if $f(k)$ is normalized (which is necessary for `ProbabilityDistribution`)
Mar 23, 2017 at 12:11	comment	added	Szabolcs		In general, you can define distributions with parameters as usual. For example `dist = ProbabilityDistribution[((-1 + x)/x) x^-k, {k, 0, Infinity, 1}]`. Here `x` is a parameter. `Mean[dist]` and `Variance[dist]` work. You will notice that the latter is a `ConditionalExpression`, because these probabilities make sense only if `x>1`. It is these kinds of assumptions that are harder to control when using `Mean`/`Variance` than when you do the `Sum`s directly.
Mar 23, 2017 at 10:37	history	edited	J. M.'s missing motivation♦		edited tags
Mar 23, 2017 at 10:12	comment	added	user44169		@Szabolcs well, I want to define it as a probability distribution to evaluate easily mean, variance and other parameters. But you right, probably is not needed at all, more like I wanted to see how to use this built-in function and if is possible to define these kind of probability distributions with various parameters.
Mar 23, 2017 at 10:09	comment	added	Szabolcs		I guess the important question is: what do you want to do with this distribution? `ProbabilityDistribution` may or may not be the best way to deal with it.
Mar 23, 2017 at 10:07	comment	added	Szabolcs		Yes, `ProbabilityDistribution` can be used when you have parameters. Just make sure that the probabilities add up to 1—it doesn't check this. Also, if you have too many parameters, many of the simple symbolic calculations (like `Mean`) may fail. Consider also setting `$Assumptions` in that case.
Mar 23, 2017 at 9:51	history	asked	user44169	CC BY-SA 3.0

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