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## Hot answers tagged martingales

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_td(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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