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May 19

Independence of Wiener process increments (expectation of product of random normal variables)

If $z_{j}=W_{t_{j}}-W_{t_{j-1}}$ and $z_{k}=W_{t_{k}}-W_{t_{k-1}}$ are two non-overlapping Wiener process increments with distribution ${\mathcal {N}}(0,t_{j}-t_{j-1})$ and ${\mathcal {N}}(0,t_{k}-t_{k-1})$ respectively, how do I prove their independence?

I know that they will be independent if they have zero covariance, and that $\operatorname {Cov} (z_{j},z_{k})=\operatorname {E} {\big [}z_{j}z_{k}{\big ]}-\operatorname {E} [z_{j}]\cdot \operatorname {E} [z_{k}].$

And I know that $\operatorname {E} [z_{j}]=\operatorname {E} [z_{k}]=0$

So I guess that all I have to show is that $\operatorname {E} {\big [}z_{j}z_{k}{\big ]}=0$

Now according to this link the product of two normal distributions has some sort of weird distribution with a Bessel function and a Dirac delta function in it, and I can't see how to take the expectation of this.

I would appreciate any advice about how to figure this out. —Preceding unsigned comment added by Thorstein90 (talk • contribs) 00:06, 19 May 2011 (UTC)[reply]

It is certainly not true that any two random variables with zero covariance are independent. For example, suppose X = −1, 0, or 1 with equal probabilities and Y = X². Then the cov(X,Y) = 0 but X and Y are obviously not independent.

But as for proving that increments of a Wiener process on non-overlapping intervals are independent: that's part of the definition. If the increments are not independent, then it's not a Wiener process. Michael Hardy (talk) 03:01, 19 May 2011 (UTC)[reply]

The formula involving the Bessel function that you linked to applies only if it is already known that the two random variables are independent. Michael Hardy (talk) 03:06, 19 May 2011 (UTC)[reply]

Conjecture about generalized Fermat numbers in base 10

How can I prove the following conjecture:

Let a be a generalized Fermat number in base 10 (in other words a number of the form 10^b+1), where b = 2c + 1 (meaning b is an odd integer, c can be any integer). Then a is divisible by 11.

I have already factored base 10 generalized Fermat numbers up to about 10¹⁷⁰+1, and the collected data seems to support this conjecture.

Any help is greatly appreciated. Toshio Yamaguchi (talk) 19:16, 19 May 2011 (UTC)[reply]

You can prove it using modular arithmetic. --COVIZAPIBETEFOKY (talk) 20:09, 19 May 2011 (UTC)[reply]

Or you can inspect the form of (10^b + 1)/11 for b = 3,5,7,9,... and this will lead you to an explicit factorisation, 10^b + 1 = 11 * something, where "something" is seen to be a whole number. 86.177.108.146 (talk) 20:22, 19 May 2011 (UTC)[reply]

Another alternative: induction on c works quite nicely too. 129.234.53.36 (talk) 20:30, 19 May 2011 (UTC)[reply]

Banach spaces and continuous linear maps

Hello everyone,

I am wondering about the following question: Suppose X, Y, Z, W are Banach spaces, and G: X x Y x W --> Z is linear and continuous in each variable separately. Does it follow G is continuous?

I have managed to prove (more or less) it is true in the case of 2 spaces X x Y, but I have not had any luck with the 3-space situation. I feel like it should not be true in this case, but I cannot construct a counterexample. Also, I am having trouble with the final step of my proof of the 2-case: I have shown that $G:X\times Y\to Z$ then $\|G(x,y)\|_{Z}\leq M\|x\|_{X}\|y\|_{Y}\forall x\in X,\,y\in Y$ , but I can't quite show (since multiplying 2 norms together does not give another norm, so it is bounded but not by an appropriate norm that I can think of) that is it bounded under a good norm to show it is continuous. However, it is the first part of the problem which is more pressing and confusing to me, so if you are only able to offer a little help I would prefer it on the X x Y x W problem!

Many thanks indeed, Mathmos6 (talk) 20:39, 19 May 2011 (UTC)[reply]

A multilinear map

G:X_{1}\times \cdots \times X_{k}\to Y

is continuous if and only if

\|G(x_{1},\dots ,x_{k})\|\leq c\|x_{1}\|\cdots \|x_{k}\|.

The proof is only a slight adjustment of the proof that a linear map is continuous if and only if it is bounded. Sławomir Biały (talk) 22:28, 19 May 2011 (UTC)[reply]

Lagrange Interpolation Method

hey all. I recently came across the Lagrange interpolation method for fitting a polynomial of degree n to n+1 points and while I'm told it's generally simpler and faster than the elementary approach (solving a system of equations) it seems like there are a lot of intermediate and indirect steps (for example finding the Lagrange polynomials). I am looking for someone with more experience than I to answer this: When is it more "worth it" ie quicker to use the Lagrange formula as opposed to solving a system? Certainly not for n=1; probably not for n=2 either; what about n=3? n=4? THanks. 72.128.95.0 (talk) 21:29, 19 May 2011 (UTC)[reply]

Did you study our article on Lagrange interpolation? The discussion there about pros and cons is OK. Bo Jacoby (talk) 09:57, 22 May 2011 (UTC).[reply]