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/sci/ - Science & Math

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6819141 No.6819141 [Reply] [Original]

Hey /sci/, I have a math problem here.

The short version:

I have a discrete distribution <span class="math">X[/spoiler], where <span class="math">P(X=n)[/spoiler] is known for all natural numbers <span class="math">n[/spoiler].

Let <span class="math">X_i = X[/spoiler] for <span class="math">i=1,2,...,m[/spoiler] - that is, we have some iids.

Define <span class="math">Z(m)=\displaystyle\sum_{i=1}^m X_i[/spoiler]. Is there an easy way to calculate <span class="math">P(Z(m)=n)[/spoiler] for all natural numbers <span class="math">m,n[/spoiler]?

>> No.6819143
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6819143

>>6819141
Here's what I find difficult:

Once m and n get big, my normal method of counting runs into partition problems.

For example, consider <span class="math">m=4[/spoiler] and <span class="math">n=6[/spoiler]. I effectively need to calculate the probability of the following outcomes, recorded as <span class="math">(X_1,X_2,X_3,X_4)[/spoiler]:

(6,0,0,0),(0,6,0,0),(0,0,6,0),(0,0,0,6)
(5,1,0,0)(5,0,1,0),(5,0,0,1),(1,5,0,0),(0,5,1,0),(0,5,0,1),(1,0,5,0),(0,1,5,0),(0,0,5,1),(1,0,0,5),(0,1,0,5),(0,0,1,5)
(4,2,0,0) x the same number of rearrangements
(4,1,1,0) x "
(3,1,1,1) x four arrangements
etc.
...

I can express it using partitions (with conditions) and multinomial coefficients, but I can't express it directly in terms of m and n. Is there some way to deal with the partitions in a clever way?

>> No.6819180

Take the Fourier Transform of the distribution of X. The Fourier Transform of the distribution of Z(m) is just the m-th power of the Fourier Transform of the distribution of X. After that calculate the inverse Fourier Transform and you have the distribution of Z(m).

>> No.6819182
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6819182

>actual science on /sci/

>> No.6819183
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6819183

>>6819143
>using a shitty casio

I found your problem pleb

>> No.6819203
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6819203

>>6819180
Hey that's really nifty.
Thanks, Anon!

>> No.6819227

For large m use the central limit theorem to approximate