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bio website jaheruddin.nl
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Dec
10
comment Expectation of constraint random walk
When I try X = [-1 -1 -1 +1 +1 +1 +1 +1 -1] your Y indeed matches my V. However when I try X = [+1 +1 +1 -1 -1 -1 -1 -1 +1] I don't see the match. Perhaps I just need to clear my mind a bit.
Dec
10
comment Expectation of constraint random walk
Thanks for your answer, but I don't see why $Y_n$ matches the dynamics of the question. For example: X = [1 1 1 -1 -1 -1 -1 -1 1] then S = [1 2 3 2 1 0 -1 -2 -1] and M = [1 2 3 3 3 3 3 3 3] so Y = [0 0 0 1 2 3 4 5 4] but based on this X the value I am interested in would be: V = [1 2 3 2 1 0 0 0 1]
Dec
10
comment Expectation of constraint random walk
@Did That was my intuition as well, however I could not find a proof or strong reasoning for this.
Dec
10
asked Expectation of constraint random walk
Dec
9
awarded  Caucus
Dec
4
comment Is it possible to simulate a floor() function with elementary arithmetic?
This is at best a supplement to an existing answer. As such I flagged this.
Oct
23
awarded  Yearling
Sep
30
awarded  Explainer
Sep
24
awarded  Autobiographer
Sep
19
comment In primary school I was showed this. Why does it work?
Why does this fail for 645? (or any A,B,C in general?)
Sep
16
comment Ambiguity of notation: $\sin(x)^2$
Your first line holds if it is only about a single variable. However, if x is defined as y+z you could see sin(y+z)^2 meaning sin((y+z)^2)
Sep
12
comment proof by induction - explanation on it
The core point here is that you choose your base case in such a way that the base case is case k for some k. (Often k=1 is chosen here)
Aug
29
revised Define an infinite subset of primes such that the sum of reciprocals converges
Without this edit the uniqueness of choices is not guaranteed. (You need to make sure to pick something below 2^n+1)
Aug
29
suggested approved edit on Define an infinite subset of primes such that the sum of reciprocals converges
Aug
29
comment Safe with $12 \times 10^6$ combinations? How is it possible?
It looks like this safe does not have an 'end of sequence' button. As such a person who wants to try a few codes of length 6, will automatically attempt a number of codes of shorter length. In any case, for reasonable safety I would recommend you to pick a code of at least length 5. Below that a patient person can crack it manually.
Aug
26
comment Average of percentage of 2 numbers.
For completeness, it should still be the weighted mean of the percentages if the prices differ. Just weighted by the sum of prices in each group.
Aug
25
awarded  Disciplined
Aug
22
comment Should I throw the dice again if I have rolled 4?
Nice note on the variability. This especially matters if you play something like 'first to five points wins and a tie counts for half'. Then you could choose to play it safe if you are close to winning the match. -- Sidenote, if only a loss gives your opponent a point as well the last line is not relevant.
Aug
22
comment Should I throw the dice again if I have rolled 4?
@TonyK I understand the answer now, if you could edit it I can correct my unjust vote.
Aug
22
comment Should I throw the dice again if I have rolled 4?
@WillieWong Thanks, it was a simple typo, though it goes against my intuition my results now agree with those by TonyK.