Re: random walk in 2d
On 12 May 2001 22:07:07 GMT, [EMAIL PROTECTED] (Francis Dermot
Sweeney) wrote:
I had got to a similar stage, and had tried showing the integral vanishes
by complex analysis methods. I had thought that since ln r is harmonic,
the value of the integral around the curve gives teh value at the center
(*2pi), but the problem is that the origin is a point on the curve where
the function fails to be harmonic.
That's exactly why that was not the argument I actually presented.
You can do it this way, but there's a detail or two. For 0r1 we have
that the integral of log(|1+re^{it}|) vanishes, and now some limiting
argument, like the Dominated Convergence Theorem if nothing more
elementary, shows that this remains true for r=1. (Quiz: What's the
value of the integral for r 1?)
Why, in your argument, does integral(log|1-exp(2it)| = I ? are n't you
traversing the same circle twice, giving twice the
integral. Maybe my complex analysis needs work.
Calculus, rather. If f(t) is a function defined on [0,2pi) and we
extend it to [0,4pi) by decreeing that f(t+2pi)=f(t) then the
integral of f(t) from 0 to 2pi is the same as the integral of
f(2t) from 0 to 2pi. (A simple change of variables shows that
the integral of f(2t) from 0 to pi = (int f(t), 0, 2pi)/2, and
similarly (int f(2t), pi, 2pi) = (int f(t), 0, 2pi)/2. Now apply the
fact that 1/2 + 1/2 = 1.)
Francis
[EMAIL PROTECTED] (David C. Ullrich) writes:
On 11 May 2001 19:17:40 GMT, [EMAIL PROTECTED] (Francis Dermot
Sweeney) wrote:
Here is a problem that is quite tricky. Starting at a radius R_o, a hop
is made of length from the current point to the origin (R_o), in a random,
uniform direction, on a 2d plane. This take us to a new point, with
distance to the
origin R_1. The next hop is then of length R_1, in a random uniform
direction, etc.
So (X_n) satisfies... oops, you use X_n for something else below.
So if H_n is the position after the n-th hop then
H_(n+1) = H_n + |H_n| * exp(i theta_n),
where the theta_n are i.i.d. uniformly distributed on [0,2 pi), right?
Or equivalently, making the notation nicer below,
H_(n+1) = H_n + H_n * exp(i theta_n)
Show that X_n=log(R_n)-log(R_(n-1)) are i.i.d. random variables, with
mean zero, and finite variance.
Well,
log(R_n) - log(R_(n-1)) = log(R_n/R_(n-1))
= log(|H_n|/|H_(n-1)|)
= log(|1+exp(i theta_(n-1))|),
which makes the iid and finite variance parts clear. To show it has
mean zero you need to show that the integral from 0 to
2 pi of log(|1+exp(it)|) equals 0. You could do that with a little
complex analysis or you could use a trick: Say this integral is I.
Then by symettry we also have I = integral from 0 to 2 pi of
log(|1-exp(it)|), and so again by symettry
I + I = integral(log(|1+exp(it)|) + log(|1-exp(it)|))
= integral(log(|1-exp(2it)|))
= I,
hence I = 0.
Francis.
--
Francis Sweeney
Dept. of Aero/Astro
Stanford U.
David C. Ullrich
*
Sometimes you can have access violations all the
time and the program still works. (Michael Caracena,
comp.lang.pascal.delphi.misc 5/1/01)
--
Francis Sweeney
Dept. of Aero/Astro
Stanford U.
David C. Ullrich
*
Sometimes you can have access violations all the
time and the program still works. (Michael Caracena,
comp.lang.pascal.delphi.misc 5/1/01)
=
Instructions for joining and leaving this list and remarks about
the problem of INAPPROPRIATE MESSAGES are available at
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Re: random walk in 2d
On 11 May 2001 19:17:40 GMT, [EMAIL PROTECTED] (Francis Dermot Sweeney) wrote: Here is a problem that is quite tricky. Starting at a radius R_o, a hop is made of length from the current point to the origin (R_o), in a random, uniform direction, on a 2d plane. This take us to a new point, with distance to the origin R_1. The next hop is then of length R_1, in a random uniform direction, etc. So (X_n) satisfies... oops, you use X_n for something else below. So if H_n is the position after the n-th hop then H_(n+1) = H_n + |H_n| * exp(i theta_n), where the theta_n are i.i.d. uniformly distributed on [0,2 pi), right? Or equivalently, making the notation nicer below, H_(n+1) = H_n + H_n * exp(i theta_n) Show that X_n=log(R_n)-log(R_(n-1)) are i.i.d. random variables, with mean zero, and finite variance. Well, log(R_n) - log(R_(n-1)) = log(R_n/R_(n-1)) = log(|H_n|/|H_(n-1)|) = log(|1+exp(i theta_(n-1))|), which makes the iid and finite variance parts clear. To show it has mean zero you need to show that the integral from 0 to 2 pi of log(|1+exp(it)|) equals 0. You could do that with a little complex analysis or you could use a trick: Say this integral is I. Then by symettry we also have I = integral from 0 to 2 pi of log(|1-exp(it)|), and so again by symettry I + I = integral(log(|1+exp(it)|) + log(|1-exp(it)|)) = integral(log(|1-exp(2it)|)) = I, hence I = 0. Francis. -- Francis Sweeney Dept. of Aero/Astro Stanford U. David C. Ullrich * Sometimes you can have access violations all the time and the program still works. (Michael Caracena, comp.lang.pascal.delphi.misc 5/1/01) = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: random walk in 2d
I had got to a similar stage, and had tried showing the integral vanishes by complex analysis methods. I had thought that since ln r is harmonic, the value of the integral around the curve gives teh value at the center (*2pi), but the problem is that the origin is a point on the curve where the function fails to be harmonic. Why, in your argument, does integral(log|1-exp(2it)| = I ? are n't you traversing the same circle twice, giving twice the integral. Maybe my complex analysis needs work. Francis [EMAIL PROTECTED] (David C. Ullrich) writes: On 11 May 2001 19:17:40 GMT, [EMAIL PROTECTED] (Francis Dermot Sweeney) wrote: Here is a problem that is quite tricky. Starting at a radius R_o, a hop is made of length from the current point to the origin (R_o), in a random, uniform direction, on a 2d plane. This take us to a new point, with distance to the origin R_1. The next hop is then of length R_1, in a random uniform direction, etc. So (X_n) satisfies... oops, you use X_n for something else below. So if H_n is the position after the n-th hop then H_(n+1) = H_n + |H_n| * exp(i theta_n), where the theta_n are i.i.d. uniformly distributed on [0,2 pi), right? Or equivalently, making the notation nicer below, H_(n+1) = H_n + H_n * exp(i theta_n) Show that X_n=log(R_n)-log(R_(n-1)) are i.i.d. random variables, with mean zero, and finite variance. Well, log(R_n) - log(R_(n-1)) = log(R_n/R_(n-1)) = log(|H_n|/|H_(n-1)|) = log(|1+exp(i theta_(n-1))|), which makes the iid and finite variance parts clear. To show it has mean zero you need to show that the integral from 0 to 2 pi of log(|1+exp(it)|) equals 0. You could do that with a little complex analysis or you could use a trick: Say this integral is I. Then by symettry we also have I = integral from 0 to 2 pi of log(|1-exp(it)|), and so again by symettry I + I = integral(log(|1+exp(it)|) + log(|1-exp(it)|)) = integral(log(|1-exp(2it)|)) = I, hence I = 0. Francis. -- Francis Sweeney Dept. of Aero/Astro Stanford U. David C. Ullrich * Sometimes you can have access violations all the time and the program still works. (Michael Caracena, comp.lang.pascal.delphi.misc 5/1/01) -- Francis Sweeney Dept. of Aero/Astro Stanford U. = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
2d random walk
Here is a problem that is quite tricky. Starting at a radius R_o, a hop is made of length from the current point to the origin (R_o), in a random, uniform direction, in 2d. This take us to a new point, with distance to the origin R_1. The next hop is then of length R_1, in a random uniform direction, etc. Show that X_n=log(R_n)-log(R_(n-1)) are i.i.d. random variables, with mean zero, and finite variance. Francis. -- __ Francis Sweeney Dept. of Aero/Astro Stanford U. = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
random walk in 2d
Here is a problem that is quite tricky. Starting at a radius R_o, a hop is made of length from the current point to the origin (R_o), in a random, uniform direction, on a 2d plane. This take us to a new point, with distance to the origin R_1. The next hop is then of length R_1, in a random uniform direction, etc. Show that X_n=log(R_n)-log(R_(n-1)) are i.i.d. random variables, with mean zero, and finite variance. Francis. -- Francis Sweeney Dept. of Aero/Astro Stanford U. = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Random Walk
In a random walk with state space = Z and transition probabilities P(k -- k+1)=p, P(k -- k)=r, P( k -- k-1)=q with p+q+r=1, the expected number of steps before moving up is either finite or infinite depending on p, q, r. This means (applied to the stock market) that it is possible for a stock price to never surpass its present value. I am looking for a proof that someone with NO mathematical background could understand. Tthe easiest proof I have so far requires knowledge of recurrence relations and basic arithmetic (+, -, *, /). I would like to put this proof on my financial web site. The author of the proof would be acknowledged. Thank you. Vincent Granville -- http://www.datashaping.com : Advanced Trading Strategies = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
