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The equation $$\text{cov} \left( \int_0^T a(t) \, dW_t, \int_0^T b(t) \, dW_t \right) = \mathbb{E} \left( \int_0^T a(t) b(t) \, dW_t \right)$$ does not hold. Instead it should read $$\text{cov} \left( \int_0^T a(t) \, dW_t, \int_0^T b(t) \, dW_t \right) = \mathbb{E} \left( \int_0^T a(t) b(t) \, \color{red}{dt} \right).$$ This follows from Itô's isometry ...

2

No: let $\Omega=\{a,b\}$, $\mathcal{F}=2^\Omega$ and suppose that your filtration is $\mathcal{F}_1=\{\emptyset,\Omega\}$ and, for any $k\ge 2$, $\mathcal{F}_k=2^\Omega$. Let $\mathbb{P}$ be given by $\mathbb{P}(\{a\})=\mathbb{P}(\{b\})=\frac{1}{2}$. Let $X_1\equiv 1$ and for $k\ge 2$ $X_k(a)=2$, $X_k(b)=0$. Then $(X_k)$ is a nonnegative converging ...

2

Hints: The stopped process $M_n := 3^{X_1+\ldots+X_{n \wedge \tau}}$ is also a martingale; hence, $$\mathbb{E}M_n = \mathbb{E}M_1.$$ Calculate $\mathbb{E}M_1$. By the strong law of large numbers, $$\frac{X_1+\ldots+X_n}{n} \to \mathbb{E}X_1 = - \frac{1}{3}.$$ Thus, $$X_1+\ldots+X_n \to - \infty \quad \text{almost surely as} \, \, n \to \infty.$$ Conclude ...

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Hint: Since $W_t=W_s+\sqrt{t-s}\,Z$ where $Z$ is standard normal and independent of $\mathscr F_s$, $$E(Z_t\mid\mathscr F_s)=\frac{G(W_s,\sqrt{t-s},1+2t)}{\sqrt{1+2t}},\qquad G(w,a,b)=E(\mathrm e^{(w+aZ)^2)/b}).$$ Hence the task is to compute the function $G$ and even, more precisely, to show that, for every $(w,a,b)$ such that $b\gt2a^2$, ...

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I don't know much about probabilities and its well-known formulas so let me try it from scratch. Assume $Z\sim N(\mu,\sigma^2)$ and let us compute $E\left[e^{Z^2}\right]$. ...

1

For the computation of $\mathbb E[M_n^2]$: the second equality sign has to be justified by the fact that the sequence $(P_j)_{j=1}^n= (X_1\dots X_j)_{j=1}^n$ is non correlated , that is, $\mathrm{Cov}(P_i,P_j)=0$ if $i\neq j$. We actually have $\mathbb E[M_n^2]=\operatorname{Var}(M_n)+(\mathbb E M_n)^2= \operatorname{Var}(M_n)$. In the proof that ...

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Here's a somewhat expanded answer, since this is a fairly straightforward question if you've covered the relevant definitions. Let $m < n$, then we have that $\mathbb{E}(X_n|I_m) = \mathbb{E}(\mathbb{E}(Y|I_n)|I_m)$. But from the definition of a filtration, we know that $I_m \subseteq I_n$, so the law of iterated expectations applies, and it follows ...

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No, it does not hold. In fact, an equivalent definition of a martingale is an integrable adapted process that satisfies the optional stopping theorem. As a counter-example, let your process be $X_n$ where $\mathbb{P}[X_i = 1] = 1/2$ and $\mathbb{P}[X_i = -1] = 1/2$ and $X_i$ are i.i.d. Let $\tau=\inf\{n: X_n =1\}$, then $\mathbb{E}[X_{\tau}]=1$. Note that ...

1

It is not difficult to see that the process $$W_t := \begin{cases} t B_{1/t}, & t > 0, \\ 0, & t=0 \end{cases}$$ defines a Brownian motion and $$\mathcal{G}_t := \sigma(B_u; u \geq t) = \sigma\left(W_u; u \leq \frac{1}{t} \right) =: \mathcal{H}_{\tilde{t}}$$ for ${\tilde{t}} := 1/t$ is the canonical filtration of $(W_\tilde{t})_{\tilde{t} \geq ... 1 Lemma: On the space$\tilde{\Omega} := \{f: [0,\infty) \times \Omega \to \mathbb{R}; f$càdlàg$\}$we define a norm by $$\|f\|_{L^1} := \mathbb{E}(\|f\|_{\infty}) := \mathbb{E} \left( \sup_{t \geq 0} |f(t)| \right).$$ Then $$L^1(\Omega; D[0,\infty)) := \{f \in \tilde{\Omega}; \|f\|_{L^1}<\infty\}$$ is a complete normed space. Proof: As a composition of ... 1 No,$T_n<\infty$does not hold true. Proof: Suppose that$T_n<\infty$almost surely. Since$(X_t)_{t \geq 0}$is a martingale,$X_0 = 1$, the optional stopping theorem yields $$\mathbb{E}X_{T_n \wedge t} = 1$$ for any$t \geq 0$. Since$|X_{T_n \wedge t}| \leq n$for all$t$, we get $$\mathbb{E}X_{T_n}=1 \tag{1}$$ by the dominated convergence ... 1 Start from the equality $$\mathbb E[|Y_{T\wedge n}|\mathbb 1\{|Y_{T\wedge n}|\gt R\} ] =\mathbb E[|Y_{T\wedge n}|\mathbb 1\{|Y_{T\wedge n}|\gt R\}\mathbb 1\{T\gt n \} ]+\mathbb E[|Y_{T\wedge n}|\mathbb 1\{|Y_{T\wedge n}|\gt R\}\mathbb 1\{T\leqslant n \} ],$$ which gives $$\mathbb E[|Y_{T\wedge n}|\mathbb 1\{|Y_{T\wedge n}|\gt R\}]\leqslant ... 1 Hints: Without loss of generality, we may assume X_0=0 (otherwise consider X_n-X_0). Since (X_n)_{n \in \mathbb{N}} is a martingale, (|X_n|^p)_{n \geq 0} is a submartingale. By the Doob decomposition, A_n \geq 0 is increasing.$$\mathbb{E}(|A_n|) = \mathbb{E}(A_n) \leq \sup_{k \in \mathbb{N}} \mathbb{E}(|X_k|^p) + \mathbb{E}(|M_0|).$$Conclude ... 1$$ Y_t = \mathrm{e}^{2B_t-\alpha t} $$using Ito$$ dY_t = -\alpha Y_t dt + 2Y_tdB_t + \frac{4}{2}Y_t dB_t^2 = -\alpha Y_t dt + 2Y_tdB_t + 2Y_t dt = (2-\alpha)Y_tdt + 2Y_tdB_t$$thus martingale requires a driftless SDE thus$\alpha = 2\$. Was this the way you performed it?

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