Recent content by MiloRambaldi

I guess you must work in finance? It is? QuantNet is a quantitative finance forum. Because finance is so far beneath you?

I guess you must work in finance? Look at the title of the thread. It invited debate on who is the best mathematician. IMHO, it the best result that counts and my impression is that none of Tao's (obviously great) achievements has equaled Perelman's achievement, no matter how many fields he...

He's not that young anymore ;) Nor the best of our time, IMHO (not quite at the level of say Perelman).

Good luck with that. Toronto is not remotely comparable to the US or London, and not just in scale.

I meant "not doable in C#" at a deeper level. In your example, you have the option of avoiding generics (e.g. MatrixDouble, MatrixInt, etc...) at the expense of code maintainability. But it seems the entire concept of arbitrary combinations of objects for the purpose of computation is not...

Interesting. Implicit inlining is relied on more than I realized.

I don't know about that. Genetic programming off the top of my head seems like it could conceivably use arbitrary levels of compositions. In fact, this caught my attention because this is what they were trying to do at the last (FT) job I had: The boss had an idea that objects representing...

You mention no run-time type erasure, but avoiding indirection can be crucial for some applications of compositions. For example, if composition is nested (n-1)-times, e.g. for n=4: auto f_4 = compose(f_3, compose(f_2, compose(f_1, f_0))); then if I am not mistaken you will have 3 levels of...

No, you plugged in z=0 on the RHS. Sorry, you need to be more careful than that ;) We can actually use the same idea as for \(p=\frac12\): Let \(g(z)=\log(f(z))'=\frac{f'(z)}{f(z)}\), which is defined on \([0,1)\) since \(f(0)>0\) as you pointed out earlier. From \(g(z)=pg(pz+q)+qg(qz+p)\), we...

To finish what we started: Given \(0<p<1\), from @zlatancrew we have \(f(z)=f(pz+q)f(qz+p)\qquad(1)\). Assume that \(E[X+Y]=f'(1)<\infty\). @zlatancrew pointed out that we have already proved it for \(p=\frac12\). Let \(\lambda=f'(1)\). As @Hi Rez suggested: Claim 1. \(f^{( n...

I think you are on the right track, and that @Hi Rez is possibly already there: If the pgf \(f\) is also entire, or at least has radius of convergence > 1, then \(f^{( k )}(1)=E[X(X-1)(X-2)...(X-k+1)] = \lambda^k\) would imply that \(f(x)=e^{\lambda(x-1)}\) because they have the same Taylor...

I can't understand why g"(1) must be defined, even for the left derivative (since 1 is not in the interior of the domain)? When we talk about the \(f^{( n )}(1)\) , this uses differentiation of the series defining f term by term.

Good point that b=-a is automatic.

I'm not sure what you mean by "the factorial moments of X are of the form \(\lambda^k\)"?

Looks perfect up to (3) but I couldn't follow the rest. It seems you were close with the idea of using the Identity Theorem. I can do the special case \(p=\frac12\) with the additional assumption that the generating function is an entire function: \(f(z)=f(\frac{z+1}2)^2\qquad\) (4) Then...