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May 5[edit]

2D Wiener Process[edit]

I have a 2D Wiener process $\mathbf {W} _{t}=(W_{t}^{1},W_{t}^{2})$ and need to find the probability that $|\mathbf {W} _{t}|<R.$

I know that each process is normally distributed with mean 0 and variance t.

So $\mathbb {P} (W_{t}^{1}<x)={\frac {1}{\sqrt {2\pi t}}}e^{-x^{2}/2t}$

and $\mathbb {P} (W_{t}^{2}<y)={\frac {1}{\sqrt {2\pi t}}}e^{-y^{2}/2t}$

and I know I need to get the joint pdf and integrate over the area where ${\sqrt {x^{2}+y^{2}}}<R$ , or transform to polar coordinates with $x=r\cos \theta$ and $y=r\sin \theta$ and integrate over a full circle and the length of the radius. I'm just unsure of where to start and what joint pdf to use. —Preceding unsigned comment added by 130.102.158.15 (talk) 06:11, 5 May 2011 (UTC)[reply]

I think it is very unlikely that there is an explicit formula for this. The joint pdf is simply the product of the two univariate pdf's, but I don't think you will be able to integrate it over a circle and get a closed form solution. When the two variances are equal it can be done using a special trick, but the trick won't work in the more general case. Looie496 (talk) 16:32, 5 May 2011 (UTC)[reply]

The above is completely mistaken; see my answer below. Michael Hardy (talk) 19:13, 5 May 2011 (UTC)[reply]

In the OP they have the same variance. Unless I'm missing something,

|\mathbf {W} _{t}|

just follows a Chi distribution. -- Meni Rosenfeld (talk) 16:55, 5 May 2011 (UTC)[reply]

Am I missing something or is this an easy problem? (W_t¹)² has a chi-square distribution with one degree of freedom, scaled so that its expected value is t. So does (W_t²)², and they are independent of each other. Therefore their sum has a chi-square distribution with two degrees of freedom, scaled so that its expected value is 2t. A chi-square distribution with two degrees of freedom is an exponential distribution.

(The original poster is mistaken in his probability statement: the expressions to the left of his two "="s are probability densities, not cumulative distribution functions.)

So

\Pr((W_{t}^{1})^{2}+(W_{t}^{2})^{2}>R^{2})=e^{-R^{2}/(2t)}.\,

Note: It if NOT TRUE that

\mathbb {P} (W_{t}^{1}<x)={\frac {1}{\sqrt {2\pi t}}}e^{-x^{2}/2t}

because the function at the end is actually the density function, not the c.d.f. Michael Hardy (talk) 19:11, 5 May 2011 (UTC)[reply]

The OP probably ment to write $\mathbb {P} (W_{t}^{1}<x)=\mathbb {P} (W_{t}^{2}<x)={\frac {1}{\sqrt {2\pi t}}}\int _{-\infty }^{x}e^{-u^{2}/2t}du$ . Bo Jacoby (talk) 07:02, 6 May 2011 (UTC).[reply]

But that's also wrong. It is the value of the Wiener process at t, not its square, that is normally distributed. Michael Hardy (talk) 13:26, 6 May 2011 (UTC)[reply]

The superskript in

W_{t}^{2}

is an index, not an exponent. Bo Jacoby (talk) 09:14, 7 May 2011 (UTC).[reply]

Use of chinese remainder theorem, probably easy question[edit]

Let n be an odd positive integer. If n is divisible by the square of a prime number p, prove that there exists an integer z such that z^p ≡ 1 (mod n) but z not ≡ 1 (mod n).

I know this uses the Chinese remainder theorem but I'm not sure how, could anyone help? I think we want to say n=mp^k with (m,p) coprime, and then find some z which is congruent to 1 modulo m, but not ≡ 1 mod p^k, but for which z^p ≡ 1 mod p^k. However, I'm not sure how to find something which even fulfils the second of these conditions, let alone the first. I think there should be a proof which doesn't use lots of other theorems (except for CRT), but I can't come up with anything. Could you help please? I suspect I'm being stupid. Sdnahzzaj (talk) 13:07, 5 May 2011 (UTC)[reply]

And another one for you which I'm struggling with! Let a be an odd integer > 1, which is not a square. Prove that there exists a positive integer n such that n ≡ 1 mod 4 and the Jacobi symbol (n/a) = -1. This is obviously a partial reverse to the result that for the jacobi symbol, a is a quadratic residue mod n then (a/n) = +1, but the converse does not necessarily hold (unlike with the Legendre symbol). If a is of the form 4k+3 then just take n=2a-1 and the proof is trivial. However, I'm struggling with the general case; WLOG supposing a=4k+1 is nonsquare, then I'm not sure what we can say: obviously every odd square is OTF 4k+1, but clearly that doesn't work in both directions. I need to use the fact that we have a non-square, but aside from saying a=(4k+1)^2 + 4m for some m, I'm not sure what we can do: could anyone help? I would prefer hints to full answers in both cases, since I'm trying to learn from this :) Sdnahzzaj (talk) 15:25, 5 May 2011 (UTC)[reply]

On the first problem, this seems like an application of Cauchy's theorem rather than the CRT. If p² divides n then p divides the order of the multiplicative group mod n, and Cauchy says there is an element of order p which what you're looking for.--RDBury (talk) 16:02, 5 May 2011 (UTC)[reply]

Ah yes, how silly of me - I guess you need Cauchy's theorem to say that the multiplicative group 'factors' as the product of prime powers so that you can use Cauchy's theorem justifiably. Thanks very much! If anyone has any thoughts on the second problem let me know :) Sdnahzzaj (talk) 16:57, 5 May 2011 (UTC)[reply]