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# 5 Appendix: Proof of Proposition 2.2

[126.1.1] The proof given below follows Ref. [30]. [126.1.2] Suppose limnμns=μs and limnμnans+bn=νs with μs and νs both nondegenerate. [126.1.3] Then it must be shown that there exist a>0 and b such that

 μ⁢s=ν⁢a⁢s+b. (182)

[126.1.4] Pick a sequence of integers n1<n2<<nk< such that limkank=a and limkbnk=b exist with 0a and -b. [126.1.5] Consider this sequence of indices from now on as fixed. [126.1.6] Then, to simplify the notation, suppose without loss of generality that limkak=a and limkbk=b.

[126.2.1] First it will be shown that 0<a<. [126.2.2] Suppose a=. [126.2.3] Let

 u=sup⁡x:lim supn→∞⁡an⁢x+bn<∞. (183)

[126.2.4] Then for v<x<u

 lim supn→∞⁡an⁢v+bn≤lim supn→∞⁡v-x⁢an+lim supn→∞⁡an⁢x+bn, (184)

and hence for every v<u it follows that νv=0 because anv+bn- with n. [126.2.5] For v>u, on the other hand, lim supanv+bn= and hence νv=1 for v>u. [126.2.6] Thus the assumption a= contradicts to νs being nondegenerate.

[126.3.1] It follows that also b must be finite. [126.3.2] In fact if limnanx+bn= then νx=1 while for limnanx+bn=- follows νx=0.

[126.4.1] Suppose now that a=0. [126.4.2] Then for every x and ε>0

 b-ε≤an⁢x+bn≤b+ε (185)

[page 127, §0]    if n is chosen sufficiently large. [127.0.1] By monotonicity of μn it follows that

 μn⁢b-ε≤μn⁢an⁢x+bn≤μn⁢b+ε. (186)

[127.0.2] If ε is chosen so that μx is continuous at the points b-ε and b+ε, then

 μ⁢b-ε≤ν⁢x≤μ⁢b+ε. (187)

[127.0.3] Because x was arbitrary it follows that μb-ε=0 and μb+ε=1. [127.0.4] Hence μx is degenerate, contrary to the conditions above.

[127.1.1] Finally, let x be such that μx is continuous at the point ax+b, and that νx is continuous at x. [127.1.2] Then

 limn→∞⁡μn⁢an⁢x+bn=ν⁢x. (188)

[127.1.3] On the other hand because limnanx+bn=ax+b one has for sufficiently large n that

 a⁢x+b-ε≤an⁢x+bn≤a⁢x+b+ε, (189)

where ε>0 is chosen such that the distribution function μ is continuous at the points ax+b-ε and ax+b+ε. [127.1.4] Hence by monotonicity

 μn⁢a⁢x+b-ε≤μn⁢an⁢x+bn≤μn⁢a⁢x+b+ε (190)

and for n

 μ⁢a⁢x+b-ε≤lim infn→∞⁡μn⁢an⁢x+bn≤lim supn→∞⁡μn⁢an⁢x+bn≤μ⁢a⁢x+b+ε. (191)

[127.1.5] Because ax+b is a point of continuity for μx and ε is arbitrary it follows that

 limn→∞⁡μn⁢an⁢x+bn=μ⁢a⁢x+b (192)

and hence νx=μax+b proving the assertion.