(Solution Library) Let (X_i) be i.i.d. taking values in -1,1,3,7,15, ... such that P(X_1=2^k-1)=(1)/(k(k+1) 2^k), k ≥q 1 (which implicitly specifies P(X_1=-1)).

Question: Let $\left(X_{i}\right)$ be i.i.d. taking values in $\{-1,1,3,7,15, \ldots\}$ such that

\[P\left(X_{1}=2^{k}-1\right)=\frac{1}{k(k+1) 2^{k}}, k \geq 1\]

(which implicitly specifies $P\left(X_{1}=-1\right)$ ).

Show $E X_{1}=0$.
Show that for all $\alpha<1$

\[P\left(S_{n}<-\frac{\alpha n}{\log _{2} n}\right) \rightarrow 1\]

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Solution: The downloadable solution consists of 2 pages
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(Solution Library) Let (X_i) be i.i.d. taking values in -1,1,3,7,15, ... such that P(X_1=2^k-1)=(1)/(k(k+1) 2^k), k ≥q 1 (which implicitly specifies P(X_1=-1)).

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