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Suppose that $X_k$ are iid $\mathrm{Bernoulli}(1/2)$for $k=1, \ldots, n$ and $$Y_n = \sum_{k=1}^n \frac{X_k}{2^k} $$
Given $U\sim \mathrm{Unif}(0,1)$ then the mgf of $U$ is $$m_U(-s) = \mathbb{E}(\exp(-sU)) = \frac{1-\exp(s)}{s} $$ (they gave it like this).

Show that $Y_n$ converges in distribution to $U$ by showing $$\lim_{n\to\infty}\mathbb{E}(-sY_n) = \frac{1-\exp(s)}{s}.$$

I computed the left hand side to be $$\mathbb{E}(-sY_n) = \frac{1}{2^n}\prod_{k=1}^n (1+\exp(-\frac{s}{2^k})) $$ and am now stuck on how to take the limit. Also, how does showing the limit converging of the MGF imply they converge in distribution?

(I'd like to do it by finding the limit of the MGF).

heropup
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hs_math_enjoyer
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  • Yes, fixed thanks! – hs_math_enjoyer Aug 24 '19 at 05:46
  • It seems like your final expression is for $m_{Y_n}(-s)=\mathbb{E}(\exp(-s Y_n))$ rather than $\mathbb{E}(-sY_n)$. I'd have instead expected $$\mathbb{E}(-sY_n)=\mathbb{E}\left(-s\sum_{k=1}^n \frac{X_k}{2^k}\right)=-s\sum_{k=1}^n\frac{1}{2^k}\mathbb{E}(X_k).$$ – Semiclassical Aug 24 '19 at 05:52
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    There are numerous duplicates of this exact question. Please search through existing answers before posting. – heropup Aug 24 '19 at 05:53

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