Cornell, BTRY 4080
Excerpt: ... n Ai . i=1 An exercise is to show that (n Ai )c = n Ac and (n Ai )c = n Ac . These are called DeMori=1 i=1 i i=1 i=1 i gan's laws. There are no restrictions on S. The collection of events, F, must be a -field, which means that it satisfies the following: (i) , S is in F; (ii) if A is in F, then Ac is in F; (iii) if A1 , A2 , . . . are in F, then Ai and Ai are in F. i=1 i=1 Typically we will take F to be all subsets of S, and so (i)-(iii) are automatically satisfied. The only times we won't have F be all subsets is for technical reasons or when we talk about conditional expectation. So now we have a space S, a -field F, and we need to talk about what a probability is. There are three axioms: (1) 0 P(E) 1 for all events E. (2) P(S) = 1. (3) If the Ei are pairwise disjoint , P( Ei ) = i=1 i=1 P(Ei ). Pairwise disjoint means that Ei Ej = unless i = j. Note that probabilities are probabilities of subsets of S, not of points of S. However it is common to write P(x) for P({x}). Intuitively, the probability ...
UNL, SHARTKE 2
Excerpt: ... or two adjacent vertices. (Hint: Study s(u) - s(v) when u and v are adjacent.) b) Determine the maximum distance between the center and the barycenter in a tree of diameter d. (Example: In the tree below, the center is {x, y}, the barycenter is {z}, and the distance between them is 1.) x y z 3. Let G be the 3-regular graph with 4m vertices formed from m pairwise disjoint kites by adding m edges to link them in a ring, as shown on the right above for m = 6. Prove that (G) = 2m8m . 4. Count the following sets of trees with vertex set [n], giving two proofs for each: one using the Prfer correspondence and one by direct counting arguments. u a) trees that have 2 leaves. b) trees that have n - 2 leaves. 5. Prove that if the Graceful Tree Conjecture is true and T is a tree with m edges, then K2m decomposes into 2m - 1 copies of T . (Hint: If every tree T with m - 1 edges is graceful, then K2m-1 has a cyclically invariant decomposition into copie ...
Berkeley, MATH 105
Excerpt: ... ficulty than I'd anticipated; I expected problem (5) to be the tricky one but on the average people had more trouble with number (4). Let's fix firmly in our minds the fact that this result is false if it is only given that 0 fj (x) A (x) for all x. Indeed consider fj = [j,j+1] and A = (with (E, A, ) = (R, B, ). Then 0 fj (x) A (x) for all x, yet fj d = 1 while |A| = 0. Another example is fj (x) = -1/j for all x R. Now all hypotheses are satisfied with A = , except that fj are not nonnegative. Again the conclusion is false. Many solutions introduced a bit of notation and then simply claimed the conclusion without offering any analysis at all. In particular, if you didn't use the monotonicity hypothesis fj fj+1 and the hypothesis that fj 0, then you can't possibly have writtenwritten a correct proof! A more sophisticated fallacy, to which several victims fell prey, is to write fj out in the form Nj fj = k=1 cj,k Bj,k where cj,k are nonnegative scalars and the sets Bj,k are pairwise disjoint for each ...