Blog

Here are some remarks
  • Is it true that if X,Y are independent, then X+Y, X-Y are independent? We will discuss this a bit in class both in discrete and continuous settings.
  • Also during discussions, the question came up whether any pair of sets A,B that are independent in the product probability space [0,1] x [0,1] must be of the form A = U x [0,1], B=[0,1] x V or be of the form A=U x [0,1], B=V x [0,1] with independent U,V in [0,1] or of the form A=[0,1] x U and B=[0,1] x V with independent U,V in [0,1]. In a continuum space, if we have positively correlated sets P[A cap B] > P[A] P[B] and negatively correlated sets P[A cap B] < P[A] P[B] then by the intermediate value theorem we can get uncorrelated sets by interpolation. If we require the deformation to be translation there can be rigidity: take any two disjoint disks A,B of the same radius in the square [0,1] x [0,1] for example. Their intersection has measure 0. By moving one to the other, we achieve A=B and P[A cap B] = P[A] greater than P[A] P[B] But if we have two congruent sets that are disjoint, P[A cap B] = 0 smaller than P[A] P[B]. By the intermediate value theorem, there is a moment during the deformation for which P[A cap B] = P[A] P[B]. Maya looked at the problem which rectangles work and it turns out that we can take arbitrary rectangles A = U x V, B = P x Q in the square [0,1] x [0,1] which can be made independent by moving them around in the square. Not all shapes work however. If A=B is the inscribed disk in [0,1] x [0,1], then we can not shift things around. Oliver looked at the combinatorial problem of which subsets of {1,2,..,n} x {1,...,n} are independent. There are M=2(n2) subsets and M(M-1)/2 possible pairs of subsets. For n=3 and n=4, no sets were found which are not of the form A = U x {1,2,...,n}, B = {1,2,...,n} x V or A= U x {1,2,...,n}, B=V x {1,2.,..,n}, where U,V are already independent in {1,2,...,n} like U={1,2}, V={2,3} in {1,2,3,4}. It looks like an interesting combinatorial problem to see how many independent pairs of subsets there are in a finite uniform probability space (Omega ={1,2,3,...,n}, 2Omega,P[A]=|A|/n).
  • We discussed once whether there are Dynkin systems (=λ-systems) that are not σ-algebras. The line of thoughts were: in a Dynkin system we can take complements and nested unions and so take arbitrary unions. With arbitrary unions we can take intersections as A intersected B is the complement of the union of the complements of A and B. The key part missed in this was cleared up by Maya: having closure under the union of a nested sequence of sets is not equivalent to have the arbitrary union. (There had been a question in class about this (I think it was Max) and I had told then that taking arbitrary union could be derived from nested unions. This is not correct however.
    Here is an example of a Dynkin system that is not a sigma algebra. It is the set of half infinite or double infinite intervals in the real line. That is I = { (a,b) or [a,b) or (a,b] where either a or b is infinite. } This is a Dynkin system as an arbitrary nested sequence of sets in this system is still in the system and because the complement of a set is in the system and because Omega=(-infty,infty) is in the system. But it is not a sigma algebra. It is not even a pi-system.
  • Here is Mathematica recursion to compute the volume of an n dimensional ball and n dimensional sphere (always radius r=1). This appeared in HW 1. What happens is that one can differentiate the formula for the ball with respect to r to get the sphere volume. This gives S(n-1) = n B(n). For example, d/dr 4 π r3/3 = 4 π r2. Then one can integrate up balls to get the sphere. For example, a circle is 2pi times the volume 1 of a 0-ball or a 2-sphere is 2pi times the volume 2 of a 1-ball. 

    B[n_]:=S[n-1]/n; S[n_]:=2Pi B[n-1]; S[0_]:=2; B[0_]:=1;


    It is simply amazing is that both the volume of n dimensional balls and spheres goes after peaking around dimension 5 (rsp 6) to zero very rapidly. Large dimensional balls are tiny! Here is the page from my notes showing this "inflation phenomenon". From Lecture 26 of Math 22 PDF:
  • Here is Mathematica code computing using brute force all the hands of the card game Z34 to see how many sets there are. I wrote this program some years ago for a History of Math course. The game of set is really nice because it is the second smallest vector space of dimension 4 over a finite field. Since it is over a field with 3 elements there is an additional algebraic structure on it which is crucial to immediately see why the answer is 1/79. So here is maybe the most dumb way to count sets. Just produce all possible hands, then all possible sets then look at the numbers .... But you have to give some credit on how elegant and poetic the code is! 
    Game=Tuples[Range[3], 4]; 
Hands=Subsets[Game,{3}]; 
M[x_,y_]:=If[x==y,x,Complement[Range[3],{x,y}]]; 
P[X_,Y_]:=Flatten[Table[M[X[[k]],Y[[k]]],{k,4}]]; 
SetQ[{X_,Y_,Z_}]:=P[X,Y]==Z; 
Sets=Select[Hands,SetQ[#] &]; 
Length[Sets]/Length[Hands]