Timeline for Expected travel of random walk in arbitrary game with multiple payouts
Current License: CC BY-SA 3.0
34 events
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| Apr 13, 2017 at 12:44 | history | edited | CommunityBot |
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| Apr 13, 2017 at 12:20 | history | edited | CommunityBot |
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| Nov 28, 2015 at 20:47 | history | edited | user147263 | CC BY-SA 3.0 |
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| Sep 29, 2015 at 12:03 | history | edited | Dan W | CC BY-SA 3.0 |
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| S Sep 24, 2015 at 18:30 | history | bounty ended | CommunityBot | ||
| S Sep 24, 2015 at 18:30 | history | notice removed | CommunityBot | ||
| Sep 19, 2015 at 6:25 | history | tweeted | twitter.com/#!/StackMath/status/645121394834075648 | ||
| Sep 18, 2015 at 20:20 | comment | added | Colm Bhandal | @mjqxxxx, you mention that the nonzero mean overwhelms the noise. I have written a new answer below elaborating on this. However, surely there is an explicit formula for the error term, given the value of the non-zero mean, and other parameters such as the variance of a step? If you have time please comment. Thank you. | |
| Sep 17, 2015 at 0:59 | comment | added | Dan W | Just to add I now understand that in the limit where N tends to infinity, $μN$ does indeed equal what I'm after. But not for a finite number such as 10000 runs, so I would like the solution for an arbitrary number of runs. Things are beginning to make a lot more sense now. | |
| Sep 16, 2015 at 21:41 | comment | added | Dan W | @mjqxxxx: Thanks that's clear, but $μN$ would then simply equal -0.025*10000 = -250. But my simulation returns around -268. Do you think my simulation is mistaken? | |
| Sep 16, 2015 at 19:13 | comment | added | mjqxxxx | Suppose the jumps $X_1,X_2,\ldots$ are i.i.d. with mean $\mu$ and variance $\sigma^2$. Let $Y_n = |X_1+X_2+\ldots+X_n|$ (the absolute distance traveled after $n$ jumps). If $\mu=0$, then $E[Y_N]\sim \sqrt{2N\sigma^2/\pi}$. If $\mu\neq 0$, then $E[Y_N]\sim \mu N$. That's it. In the $\mu\neq 0$ case, the nonzero mean overwhelms the noise. | |
| Sep 16, 2015 at 19:05 | answer | added | Colm Bhandal | timeline score: 2 | |
| Sep 16, 2015 at 18:30 | history | edited | Dan W | CC BY-SA 3.0 |
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| S Sep 16, 2015 at 16:45 | history | bounty started | Dan W | ||
| S Sep 16, 2015 at 16:45 | history | notice added | Dan W | Draw attention | |
| Sep 16, 2015 at 16:44 | comment | added | Dan W | @Did: Thanks. Regardless, that still leaves a lot to the imagination with my current level of mathematical knowledge, but I'll open a bounty. | |
| Sep 16, 2015 at 16:43 | history | edited | Dan W | CC BY-SA 3.0 |
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| Sep 16, 2015 at 5:34 | comment | added | Did | Thiis is becoming repetitive, but... First, your (0.75∗−1+0.1∗0.5+0.15∗4.5) should read (0.75∗−1+0.1∗0.5+0.15∗4.5)*10000. Second, asymptotics are a form of information about, say, a sequence, when $N\to\infty$ only, hence a term like o(10000) is just absurd. | |
| Sep 16, 2015 at 4:31 | comment | added | DirkGently | Even the first result that you cite has smaller error terms since it relies on the central limit theorem. | |
| Sep 16, 2015 at 4:09 | history | edited | Dan W | CC BY-SA 3.0 |
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| Sep 16, 2015 at 2:07 | comment | added | Dan W | @Did: Had time to revisit this. Assuming N=10000, I'm not sure how to resolve $(0.75*-1 + 0.1*0.5 + 0.15*4.5) + o(10000)$ to equal -268.8. The first part of that sum equals -0.025. The second part (o(N)) from what I understand is the set of functions which have a lower rate of growth than N, so it could be almost anything. Somehow, I'm supposed to ascertain around -269 from that? Perhaps I could open a bounty if the expansion is more complex than I anticipated. | |
| Aug 17, 2015 at 17:42 | comment | added | Did | In Landau's notation, o(N) is any term negligible with respect to N when N goes to infinity. | |
| Aug 17, 2015 at 15:25 | comment | added | Dan W |
@Did: Can you expand the formula for clarity? I think by a 'mean increment', you're referring to the expected payout (which is $prob_1*pay_1 + prob_2*pay_2 + prob_3*pay_3$), but I'm not sure how to interpret o(N) from your comment...
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| Aug 17, 2015 at 15:11 | comment | added | Did | Sure: if the increments have mean $\mu$, the displacement after $N$ steps is $\mu N+o(N)$ (hence your 268.8 is probably minus 268.8). | |
| Aug 17, 2015 at 15:07 | comment | added | Dan W | @Did: Ah, in that case, can you give me the new revised formula which takes into account non-centered paytables? | |
| Aug 17, 2015 at 15:05 | comment | added | Did | Your example is not centered, thus the asymptotics scales like $N$ instead of $\sqrt{N}$, and CLT and variance become irrelevant. | |
| Aug 17, 2015 at 14:27 | comment | added | Dan W | @Did: Thanks for the feedback. Your formula works for the coin and dice rolls, but when I tried for three entries or more, it differs from the simulation. An example is: probabilities = [0.75, 0.1, 0.15] and payout = [-1, 0.5, 4.5]. That's a variance of 3.811875, but the average 'travel' according to your formula is 155.779 (to 3dp), whilst I get approx 268.8. Could you check your formula? | |
| Aug 17, 2015 at 11:29 | comment | added | Did | But you have already been explained that, no? | |
| Aug 17, 2015 at 11:26 | comment | added | Did | In full generality, the asymptotics is $\sqrt{2\sigma^2N/\pi}$ where $\sigma^2$ is the variance of a single step. In your dice example, $\sigma^2=5$ hence, for $N=10'000$, $\sqrt{2\sigma^2N/\pi}=100\sqrt{10/\pi}=178.4124116...$ | |
| Aug 17, 2015 at 11:06 | history | edited | Dan W | CC BY-SA 3.0 |
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| Aug 17, 2015 at 10:18 | history | edited | Dan W | CC BY-SA 3.0 |
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| Aug 17, 2015 at 9:46 | history | edited | Dan W | CC BY-SA 3.0 |
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| Aug 17, 2015 at 9:05 | history | edited | Dan W | CC BY-SA 3.0 |
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| Aug 17, 2015 at 8:53 | history | asked | Dan W | CC BY-SA 3.0 |