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Define a martingale by $$M_t := \begin{cases} \mathbb{E}(\phi(W_T) \mid \mathcal{F}_t), & t \leq T, \\ \phi(W_T), & t>T. \end{cases}$$ Then, by the martingale representation theorem, there exists a representation of the form $$M_t = c+ \int_0^t \beta_s \, dW_s.$$ For $t=T$, this proves the claim. Remark: One possibility to prove the mentioned ...

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Obviously, it only works for large $n$. Heuristically, the explanation is that although $X_n$ and $Y_n$ are not actually independent, after a large number of steps $n$, the information about the individual horizontal components of $X_n$ is essentially lost, so that little can be deduced about the distribution of $Y_n$, except that $n$ and $X_n$ together ...

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Ok, I think I've got it: We want to show that $\Pr ( S_n - (1-\epsilon)n \geq M) \to 1$, hence it's the same as proving that $\Pr ( S_n - (1-\epsilon)n < M) \to 0$. I'll use Chebyshev's inequality: $$\Pr(|X-\mathrm{E}X|\leq \epsilon) \leq \frac{\mathrm{Var}X}{\epsilon^2}$$ The key to solve this problem was to see that for $n$ large enough: $$\Pr(S_n - ... 1 If the segment is horizontal and its length equals R, the allowed positions for its centre are given by the region depicted below: where the area of the upper and lower circular segment together is given by$$\frac{\pi}{3}R^2-\frac{\sqrt{3}}{2}R^2, so the area of the striped region is given by the area of the circle, minus the previous area, minus the ...

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fix $\delta > 0$ $P(\sqrt{n}|X_n-Y_n|>\delta)$ = $P(|X_n-Y_n|>\frac {\delta}{\sqrt{n}})$ ≤ $\frac n{\delta^2} E||X_n-Y_n||^2$ = $\frac {o(1)}{\delta^2}$ < $\epsilon*1$ = $\epsilon$ for large enough N := $N_{\epsilon}$. $\Rightarrow \sqrt{n}(X_n-Y_n)$ = $o_p(1)$ Because $P(\sqrt{n}|X_n-Y_n|>\delta)$ can be made arbitrarily small for all ...

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For a fixed $A\in B_\infty$, consider $\mathcal L_A:=\{B: m(A\cap B)=m(A)m(B)\}$. It suffices to show that each $\mathcal L_A$ is a $\lambda$ system, that is, $L\in\mathcal L_A$ implies that $L^c\in\mathcal L_A$ and if $L_n\in \mathcal L_A$ for each $n$, and $L_n\subset L_{n+1}$ (which is missing in the OP), then \$\bigcup_{n\geqslant 1}L_n\in \mathcal ...

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