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2

Here is a proof that $\lim_{n\rightarrow\infty} q_n = \sqrt{2} -1$ with probability 1: Let $\liminf q_n$ and $\limsup q_n$ represent the random variables $\liminf_{n\rightarrow\infty} q_n$ and $\limsup_{n\rightarrow\infty} q_n$, respectively. Let $\{a_k\}_{k=0}^{\infty}$ be the deterministic sequence that satisfies $a_0=1/2$ and: $$a_{k+1} = \frac{1}{2}-\...


2

$dX_t = (\beta-\alpha X_t) \, dt + \sigma \, dB_t$ Forget a moment about the SDE and consider the associated ordinary differential equation $$dx_t = (\beta- \alpha x_t) \, dt \tag{1}$$ instead. If I would ask you to solve this ODE, you would (hopefully) first solve the homogeneous equation $$dx_t = -\alpha x_t \, dt$$ and find that the solution of this ...


1

By the $L^1$-convergence, we have $$L^1-\lim_{s \to \infty} \mathbb{E}(X(t+s) \mid \mathcal{F}_t) = \mathbb{E} \left( X(\infty) \mid \mathcal{F}_t \right). \tag{1}$$ On the other hand, as $(X(t))_{t \geq 0}$ is a martingale, we have $$\mathbb{E}(X(t+s) \mid \mathcal{F}_t) = X(t)$$ for all $s \geq 0$. Hence, trivially, $$L^1-\lim_{s \to \infty} \mathbb{E}...


1

Note that $\sup_t E[|X_{t\wedge n}; |X_{t\wedge n}|\ge M]=\sup_{t\in [0,n]} E[|X_{t}|; |X_{t}|\ge M]<\infty$ (left-limits existing implies bounded on compact intervals [need to verify; know it's true if left and right limits exist]). Your reasoning doesn't work. If each sample path $[0,n] \ni t \mapsto X_t(\omega)$, this does not imply $$\sup_{t \in [0,...


1

This is discrete time, so right continuity is not an issue. It is simply the fact that if $P[D_n]=1$ for all $n$ then $P\left[\cap_n D_n\right]=1$ (and conversely).


1

For 5.: The key is this: A previsible martingale is constant in time. In more detail, if $X$ admits a second decomposition $X_n=X_0+M'_n+A'_n$ then by subtracting you find that $M_n-M'_n$($=A'_n-A_n$) is both a martingale and previsible, hence constant in time a.s., whence $M_n-M'_n=M_0-M'_0=0-0=0$ for all $n$, a.s.


1

$\{S(k)\le m\} =\{A_{m+1}>k\}\in\mathscr F_m$, because $A$ is previsible.


1

Let $ X_t=aB(t)-t$ and $Y_t=\exp(2B_t-2t)$, we have $$dX_t=-dt+adB_t$$ and $$dY_t=\underbrace{\left(-2e^{2B_t-2t}+\frac{1}{2}(2)^2e^{2B_t-2t}\right)}_{0}dt+2\,e^{2B_t-2t}\,dB_t=2\,e^{2B_t-2t}\,dB_t$$ $$d(X_tY_t)=Y_tdX_t+X_tdY_t+d[X_t,Y_t]$$ as a result $$d(X_tY_t)=e^{2B_t-2t}(-dt+adB_t)+(aB(t)-t)2e^{2B_t-2t}\,dB_t+2a\,e^{2B_t-2t}dt$$ thus $$d(X_tY_t)=(-1+2a)...



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