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May 8, 2019 at 8:58 comment added Dieter Kadelka @MattF. First I computed $N = 1000000$ realisations $(z_k)_{k=1}^N$ of $\sum_{i=1}^{100} Z_i$ and saw that $z_k < 0$ for each $k$. To exclude the possibitlty of a fault in my ad hoc program I counted the number of $z_k > -30$. This gave the estimation $P(Z > -30) \sim 0.00009$.Even this extimation is much smaller than $\Phi(-10/3)$ and $f(100)$ is even much smaller. Of course, having the values $z_1,\ldots,z_N$ it is possible to calculate a kernel density estimator of $\sum_{i=1}^{100} Z_i$, f.i. with R.
May 7, 2019 at 16:28 vote accept Bogdan
May 7, 2019 at 1:26 comment added user44143 @DieterKadelka, why are you looking at $P(Z<-30)$ to calculate $f(100)$?
May 7, 2019 at 1:07 answer added Iosif Pinelis timeline score: 1
May 4, 2019 at 13:02 comment added Dieter Kadelka Sorry, the probability is $P(Z > -30) \sim 0.00009$.
May 4, 2019 at 9:47 comment added Dieter Kadelka @Bogdan: I have simulated the problem for myself. The estimation of Matt F. seems to be useless for your problem. For instance with 1000000 runs I got an estimation of the probability $P(Z < -30) = 90/1000000 = 0.00009$, far away from $0.000429$. Confirming your obeservation, even with 1000000 runs there was no success for the original problem.
May 4, 2019 at 8:17 comment added Dieter Kadelka @Bogdan: The estimation given by Matt F. seems to be very crude. The same sort of problems arise in estimating probabilities in risk theory. The technique of "Importance Sampling" (en.wikipedia.org/wiki/Importance_sampling) may be useful in your situation.
May 3, 2019 at 18:00 comment added user44143 $Z$ has mean $-1$, standard deviation $3$, so $f(100) \sim \Phi(-10/3) \sim .000429$.
May 3, 2019 at 16:38 comment added user44143 I edited the title to be consistent with the post.
May 3, 2019 at 16:37 history edited user44143 CC BY-SA 4.0
edited title
May 3, 2019 at 16:32 history asked Bogdan CC BY-SA 4.0