> From: Brian Schott <[EMAIL PROTECTED]>
>
> The poisson distribution is almost identical to the
> normal distribution when its parameter is 15 or larger, so
> you may want to use the normal distribution. More
> specifically, according to George Fishman, as the mean,
> lambda, increases the distribution of
> (X-lambda)/sqrt(lambda) converges to the standard normal
> distribution (ie mean 0, std dev 1); if Y is from standard
> normal then
>
> X=max(0,integer [lambda+Y*sqrt(lambda)+0.5])
>
> A key feature would be to generate the poisson
> cumulative probabilities only one time as you did in your
> revised approach, whether you use the Poisson or the normal
> approximation. If the mean of the Poisson does not vary,
> then why not generate all 1 000 000 at once?
>
> It seems to me that you could generate 1000000
> Poisson and uniform variates all at one time (in an array of
> shape 500 2000?) and then focus on keeping track of the
> (ending) inventory level (i) in your main process using J's
> strength of array processing. This is the challenging part,
> imo, especially avoiding the for. for iteration.
dstat poissonrand 100 10000
sample size: 10000
minimum: 64
maximum: 141
median: 100
mean: 100.006
std devn: 10.0345
skewness: 0.12287
kurtosis: 3.03206
dstat <.100+10*normalrand 10000
sample size: 10000
minimum: 61
maximum: 137
median: 99
mean: 99.3547
std devn: 10.0235
skewness: 0.0298283
kurtosis: 3.0363
plot (i.~ #/. ]) /:~ poissonrand 100 10000
plot (i.~ #/. ]) /:~ <.100+10*normalrand 10000
3 ts '$<.100+10*normalrand 10000'
0.00326381 1.31347e6
3 ts '$poissonrand 100 10000'
0.513158 690880
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