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Similar question has been discussed previously; see Why does the short rate in the Hull White model follow a normal distribution?. Basically, the probabilistic limit of normal random variables is still normal. Then, as $$\sum_{[t_{i-1},t_{i}]\in\pi_{n}}f(t_{i-1})(W_{t_{i}}-W_{t_{i-1}})$$ is normal, the limit $$\int_{0}^{t}f(\tau)dW_{\tau},$$ in probability, ...

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Note that $A$ and $B$ are not independent; when you scaling $B$, the volatility of $A$ is also scaled. That is, it is better to denote by $\sigma_{AB}$ for $\sigma_A$, to highlight the dependence on $B$. Then, If $\sigma_B > \sigma_{AB}$, you consider the volatility $B_t$ to be more volatile than $A_t$. After the scaling, you have that ...

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When you scale your process you set the following: $\tilde{B}_t=f(B_t)=\sigma_A^2 B_t$ so then by means of Ito-Lemma, you get, $df(B_t)=\partial_x f(B_t)dB_t+\frac{1}{2}\partial_{x^2}f(B_t)dt=\sigma_A^2 dB_t$ hence $d\tilde{B}_t=\sigma_A^2dB_t=\sigma_A^2 \kappa(\hat{B}_t-B_T)dt+\sigma_A^2\sigma_B dW_t^B$ but you can still symplify it and nothing has ...

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Note that, for $0 \leq s < t$, \begin{align*} W_t^3 &= (W_t-W_s+W_s)^3\\ &= (W_t-W_s)^3 + 3(W_t-W_s)^2 W_s + 3 (W_t-W_s) W_s^2 + W_s^3. \end{align*} Moreover, \begin{align*} E\big( (W_t-W_s)^3 \mid \mathcal{F}_s\big) &= E\big( (W_t-W_s)^3\big)\\ &= 0,\\ E\big((W_t-W_s)^2 W_s \mid \mathcal{F}_s\big) &= W_s E\big( (W_t-W_s)^2\big)\\ ...

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You know that $E\left[\int_{0}^{s}W_udu\right]=E\left[\int_{0}^{t}W_vdv\right]=0$. By definition \begin{align} & Cov\left(\int_{0}^{s}W_u\,du\,\,,\,\int_{0}^{t}W_v\,dv\right)=E\left[\int_{0}^{s}W_u\,du\int_{0}^{t}W_v\,dv\right]-0 \end{align} then \begin{align} & ...

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Let $S_t$ and $B_t$ be respectively the stock price and the money market account value at time $t$. Then $S_t/B_t$ is called the discounted stock price. Note that \begin{align*} E\left(\frac{S_N}{S_0}\right) &= E\left(\frac{S_N}{B_N} \frac{B_N}{B_0}\right)\frac{B_0}{S_0}\\ &= E\left(\frac{S_N}{B_N}\right) E\left(\frac{B_N}{B_0}\right)\frac{B_0}{S_0} ...

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In this case it is just the notion that your payoff function should not explode at some point - made mathematically rigorous. Have a look at the following picture from wikipedia: Intuitively the Lipschitz condition (or Lipschitz continuity) ensures that your payoff function always remains entirely outside the white cone, so it cannot e.g. become ...

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This is the Black Scholes Call Price: \begin{align} C(S, t) &= N(d_1)S - N(d_2) Ke^{-r(T - t)} \\ d_1 &= \frac{1}{\sigma\sqrt{T - t}}\left[\ln\left(\frac{S}{K}\right) + \left(r + \frac{\sigma^2}{2}\right)(T - t)\right] \\ d_2 &= \frac{1}{\sigma\sqrt{T - t}}\left[\ln\left(\frac{S}{K}\right) + \left(r - \frac{\sigma^2}{2}\right)(T - ...

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Ito integrals differ in their complexity depending on which function you'd like to integrate. Let's talk about the simplest but still powerful version - the one mentioned by Innombrabre, for integrands from the $L^2$ space. In general, you can think of the procedure as follows. You have two complete metric spaces $(X,d_X)$ and $(Y,d_Y)$. Let $A$ be a subset ...

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