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1 Nov 2018: XN m for all N n) = 1. Suppose the transition probabilities satisfy instead. ... 0 i < b, where 0 < λi, µi < 1for all i, and λi µi = 1 for 1 i < b.

16 Oct 2018: znσn and B(z) =n0. znbn. 1. Show that χ(z) z1 exp (2B(z) 2) for all z 0. ... 4. Show that B((1 )/κ) 1. 5. Conclude that there exists a positive constant c so that σn exp (cn) κn for all n 0.

16 Oct 2018: 4. Consider simple random walk on the binary tree of depth k with n = 2k1 1 vertices(the root has degree two and all other nodes except for the leaves have ... 14. Let G be a locally finite graph. Show that all trees of the free and the wired

9 Dec 2018: Broken lines represent MFI values for each. sorted B-cell subset. (B) Principal component analysis of the MFI values for each sorted cell in all samples. ... Each B-cellsubset signature contained 27 to 54 genes, ensuring compa-rable gene representation

29 Nov 2018: Rdg(x)fn(x) dx Rd g(x)f (x) dx for all bounded. continuous functions g : Rd R. ... all estimators of f0 based on X1,. , Xn. Then foreach d N, there exists cd > 0 such that.

3 Oct 2018: Then as n we get. Pp(N = k, all k -clusters intersect B(n)) 1. ... Pp(all -clusters intersect B(n)) Pp(all edges of B(n) are open) 1.

21 Nov 2018: Since Wk10 Wk0 , for all M and all n sufficiently large, we have. ... Observe that this trivially holds (for all n large enough) if µt 3.

4 Jan 2018: Therefore, Jv = 0 for all v V. Suppose ju1,u2 > 0 forsome edge 〈u1, u2〉. ... A 0/flow is a vector j = ( ju,v : u,v V, u 6= v) satisfying (1.14)(a),(b) and also (c) for all u 6= 0.