# Tag Info

3

Hi this is not always possible and it is the subject of one famous theorem in financial mathematics known as the Fundamental Theorem of Asset Pricing that claims that under some conditions there exist such measure change that turns (morally) semimartingales into local martingales. In a series of articles by Delbean and Schachermayer (available on their ...

3

Note that $T_n=uX_n - \frac12nu^2\sigma^2$ defines a random walk $(T_n)$ whose steps have mean $E(uY_1- \frac12u^2\sigma^2)=- \frac12u^2\sigma^2\lt0$ if $u\ne0$ hence, for every $u\ne0$, $T_n\to-\infty$ almost surely, $Z^u_\infty=0$ almost surely and $Z^u_n\ne E(Z^u_\infty\mid\mathcal F_n)$. The case $u=0$ is direct.

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Note that $E[X \mid \sigma(Y)]$ is $\sigma(Y)$-measurable. We will prove the existience of a measurable $h$ for any $\sigma(Y)$-measurable random variable $Z$. First let $Z = 1_A$ be a characteristic function of some $A \in \sigma(Y)$. Then $A = Y^{-1}[B]$ for some Borel set $B$. Let $h = 1_B$, then for any $\omega \in \Omega$: $$1_B\circ Y(\omega) = 1 \iff ... 2 Note that (M_t)_{t \geq 0} has independent stationary increments and its mean equals$$\mathbb{E}M_t = \mathbb{E}N_t - \lambda t = \lambda t - \lambda t = 0$$since (N_t)_{t \geq 0} is a Poisson process. Hence,$$\begin{align*} \mathbb{E}(M_t^2 \mid \mathcal{F}_s) &= \mathbb{E}((M_t-M_s)^2 \mid \mathcal{F}_s) + 2 M_s \mathbb{E}(M_t-M_s \mid ...

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Recall that $$\mathbb{E}e^{c B_t} = e^{\frac{1}{2} c^2 t} \tag{1}$$ as $B_t$ is Gaussian with mean $0$ and variance $t$. In particular, we see that $$M_t := 4^{B_t} = \exp \bigg( B_t \cdot \log 4 \bigg)$$ is not a martingale since $$\mathbb{E}M_t \stackrel{(1)}{=} \exp \left( \frac{1}{2} (\log 4)^2 \cdot t \right)$$ is not constant. In fact, by the ...

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Hi your first intuition is correct. Formally you could write for example to hsow the statement that for $t>s>1$ : $E[W_t | \mathcal{F}_s]\not=W_s$ For your second question it is more a direct application of your course if you want my opinion. For $t<1$ Have you seen what the Quadratic variation a Brownian motion is ? For t>1, you can check ...

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Using Ito's Lemma $d(2 W_t^3 + \beta t W_t) = 6 W_t^2 dW_t + \beta t dW_t + (6 W_t dt + \beta W_t dt)$ The first two terms above don't matter because they will be martingales if you take the integral of both sides. The term in parenthesis is what you want to make $0$. Therefore of course $\beta = -6$ This is actually also consistent with the method you ...

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Ok, so to show that $E[M |F_T] = M_T$ we have that letting $B \in \mathcal{F}_T$ and $A_i = B \cap \{T = i\}$ that $E[M_T 1_B] = \sum_i E[M_i 1_{A_i}] = \sum_i E[\, E[M | \mathcal{F}_i] 1_{A_i}] = \sum_i E[M 1_{A_i}]= E[M1_B]$. Now, it is clear that $E[M|F_T] = M_T$ and the above works fine.

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Question 1: Yes, that's it - because the Brownian motion $(W_t)_{t \geq 0}$ is a martingale. Question 2: Fix $i<k$. By the definition of $I_t^n$ we have \begin{align*} I_t^n &= \sum_{j=0}^{i-1} \Delta_{t_j}(W_{t_{j+1}}-W_{t_j}) + \sum_{j=i}^{k-1} \Delta_{t_j}(W_{t_{j+1}}-W_{t_j}) + \Delta_{t_k}(W_t-W_{t_k}) \\ &=: J_1+J_2+J_3 \end{align*} ...

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$$M_{t+h} - M_t = \int_t^{t+h} (1+u^2)dW_u = \lim \sum_j (1 + t_{j-1}^2) (W_{t_j} - W_{t_{j-1}})$$the limit being in $L^2$, as $\sup [t_j - t_{j-1}] \to 0$. Now, let $X$ be a random $L^2$ variable, mesurable with respect to the $F_t$ filtration. As $\sum_j (1 + t_{j-1})(W_{t_j} - W_{t_{j-1}})$ depends on the increments of $W$ afer $t$, then $$E ... 1 Without loss of generality, we may assume that the filtration (\mathcal{F}_t)_{t \in [0,T]} is right-continuous, i.e. \mathcal{F}_{t+} = \mathcal{F}_t (see the lemma below). By the backward martingale convergence theorem, this implies in particular$$\mathbb{E}(U \mid \mathcal{F}_s) = \mathbb{E}(U \mid \mathcal{F}_{s+}) = \lim_{r \downarrow s} ...

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If $M$ is a martingale and $X$ is properly integrable, then $$I_t := \int_s^t X(r) \, dM_r, \qquad t \geq s$$ defines a martingale and, consequently, $$\mathbb{E} \left( \int_s^t X(r) \, dM_r \mid \mathcal{F}_s \right) = \mathbb{E}(I_t \mid \mathcal{F}_s) = I_s =0$$ for any $t \geq s$. Now, if $M$ is a local martingale and $X$ locally integrable, then ...

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