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25

This type of integral has appeared so many times and in so many places; for example, here, here and here. Basically, for each sample $\omega$, we can treat $\int_0^t W_s ds$ as a Riemann integral. Moreover, note that \begin{align*} d(tW_t) = W_t dt + tdW_t. \end{align*} Therefore, \begin{align*} \int_0^t W_s ds &= tW_t -\int_0^t sdW_s \tag{1}\\ &= \...


17

These are all examples on Ito Formula in its general form (with quadratic variations):


13

Baxter and Rennie say it better than me, so I will summarize them. Suppose that $N_t$ is not stochastic and $f(.)$ is a smooth function then the Taylor expansion is $$ df(N_t) = f'(N_t)dN_t + \frac{1}{2}f''(N_t)(dN_t)^2 + \frac{1}{3!} f'''(N_t)(dN_t)^3 + \ldots $$ and the term $(dN_T)^2$ and higher terms are zero. Ito showed that this is not the case in the ...


11

My understanding is because the Ito's integration definition keeps the martingale property. With Brownian motion $W(t, \omega)$ defined, to define stochastic integration in a Riemann–Stieltjes style: $$\int_0^t f(t, \omega) d W(t, \omega) = \lim_{\| \Delta_n\| \to 0 } \sum_{i=1}^{n} f(\tau_i,\omega) \left ( W(t_i, \omega) - W(t_{i-1}, \omega) \right ) $$ , ...


11

From $(2)$, \begin{align*} \ln S_t &=\ln F_{t, t} \\ &= \ln F_{0, t}-\frac{1}{2}\int_0^t\sigma^2 e^{-2\lambda (t-s)}ds+\int_0^t \sigma e^{-\lambda(t-s)} dB_s\\ &=\ln F_{0, t}-\frac{\sigma^2}{4\lambda} \left(1-e^{-2\lambda t}\right)+e^{-\lambda t}\int_0^t \sigma e^{\lambda s} dB_s. \end{align*} Then, \begin{align*} \lambda e^{-\lambda t}\int_0^t \...


10

I think this question has no easy answer but I'll give it a shot anyway (beware: oversimplification ahead!). The main idea of the Malliavin calculus is to be able to differentiate stochastic processes like Brownian motion (or more general martingales with bounded quadratic variation), which are not differentiable in the traditional sense (because of their ...


10

I will try to answer this a bit differently. The rigorous answer: because Ito calculus tells us that we need the second order term. Look at $$ S_t = S_0\exp(\mu t + \sigma B_t). $$ Assume that $S_0$ is known and fixed and look at by Ito's formula $$ d(S_t/S_0) = \mu dt + \sigma B_t + \frac{\sigma^2}{2} dt. $$ Then with some abuse of notation: $$ E[d(S_t/...


10

For any $s \geq t$, note that \begin{align*} r_s = r_t + \sigma\int_t^s dW_u + \int_t^s \theta_u du. \end{align*} Then, \begin{align*} \int_t^T r_s ds &= (T-t)r_t + \sigma\int_t^T\int_t^s dW_u ds + \int_t^T \int_t^s\theta_u du ds\\ &=(T-t)r_t + \sigma\int_t^T\int_u^T ds\, dW_u +\int_t^T\int_u^T\theta_u ds du\\ &=(T-t)r_t + \sigma\int_t^T (T-u)...


9

In fact Ito and Stratonovich calculus are both mathematically equivalent. In the following paper you can e.g. see that both derivations lead to the same result, i.e. the Black-Scholes equation: Black-Scholes option pricing within Ito and Stratonovich conventions by J. Perello, J. M. Porra, M. Montero and J. Masoliver From the abstract: Options financial ...


9

Shreve's theorem also called "Girsanov II" indeed represents a special case of the general "Girsanov I" from Wiki above, with $$Y_t:=W_t,$$$$X_t:=-\int_0^t\Theta_udW_u$$ We can show: $$[Y,X]=-\int_0^t\Theta_udu$$ by using general Stochastic Calculus rules (e.g. p.37, 6.6 here): $$[Y,X]=[W_t,-\int_0^t\Theta_udW_u]=-\int_0^t\Theta_ud[W_u,W_u]=-\int_0^t\...


9

In the integral $$\int_0^t S_u dW^{*}_u \, ,$$ $dW^{*}_u \equiv W^{*}_{u+du} - W^{*}_u$ is independent from the integrand $S_u$. So, $\mathbb{E}\left[ \int_0^t S_u dW^{*}_u\middle\vert \mathcal{F}_0\right] = \int_0^t \mathbb{E}\left[S_u \middle\vert \mathcal{F}_0\right]\mathbb{E}\left[dW^{*}_u\middle\vert \mathcal{F}_0\right] = 0$, since $\mathbb{E}\...


8

Okay so I'll take Jase answer and format it properly so that it answers your question and it will be useful for users in the future. For clarity, let me restate the dynamics of the Modified Ornstein-Uhlenbeck model using the more common notation: $$dS_t = \theta (\mu-S_t)dt + \sigma S_t dW_t$$ This blog post provides a closed form solution: $$ S_t = S_0 \...


8

In general, if you have a process that you can write under the form $F(B_t,t)$ where $F$ is $\mathcal{C}^{2,1}$ then Itô's lemma gives you the drift term and diffusion term of $dF$. Then if the resulting SDE has a null drift (that's where Black Scholes PDE comes from), and you get a only local martingale. For it to be a proper martingale you can look at ...


8

The convexity of the exponential function of the stochastic variable $W$ makes its expectation greater than the exponentiation of the expectation of $W$. This is an example of Jensen's inequality, $E[e^{\sigma W}]> e^{\sigma E[W]}=1$. $\sigma$ can be interpreted as the magnitude of the convexity of the exponential function. This can be seen by Taylor ...


8

We know that $(\tilde{W}_t) := (-W_t)$ is also a Wiener process so $$ E[W_pW_qW_r] = E[\tilde{W}_p\tilde{W}_q\tilde{W}_r] = (-1)^3E[W_pW_qW_r] $$ and that implies that $E[W_pW_qW_r] = 0$.


8

Because you can hedge. Once you have delta hedged, the pay-off is symmetric about up and down moves so drift doesn't matter. Also the delta-hedged call and the delta hedged put have to have the same value since they have the same pay-off. (Put-call parity) Yet any argument that the call should be worth more because of drift says that the put should be ...


8

A stochastic differential equation is nothing more than a short-hand notation for a corresponding integral equation. So the initial SDE you provided actually means $$ \int_0^t d S_u = \int_0^t \mu(S_u, u) du + \int_0^t\sigma(S_u, u) dW_u$$ This is how the SDE is defined (see e.g. here). The reason is that you cannot differentiate a Brownian motion. It does ...


8

Just to add to the already nice answers, the result can also be obtained using the (stochastic) Fubini theorem. \begin{align} \int_0^t W_s ds &= \int_0^t \int_0^s dW_u\, ds \tag{$W_s=\int_0^s dW_u$}\\ &= \int_0^t \int_u^t ds\,dW_u \tag{Fubini} \\ &= \int_0^t (t-u) dW_u \tag{$\int_u^t ds = t-u $} \end{align} And we fall back on the same equation ...


8

Let $$Z_t = \exp(-X_t)$$ with $$X_t = \sigma(T-t)W_t+\sigma\int_0^tW_sds+\int_0^Tf(0,u)du+\int_0^t\int_s^T\alpha(s,u)du ds $$ and $W_t$ a standard Brownian motion, along with the usual assumptions. We can write $X_t=f(t,W_t)$ and apply Itô's lemma to get: $$ dX_t = \frac{\partial f}{\partial t}(t,W_t) dt + \frac{\partial f}{\partial W_t} (t,W_t) dW_t + \...


8

Note that the Ito integral of a deterministic integrand $f: \mathbb{R}_+ \rightarrow \mathbb{R}$ is normally distributed \begin{equation} \int_0^t f(u) \mathrm{d}W_u \sim \mathcal{N} \left( 0, \int_0^t f^2(u) \mathrm{d}u \right). \end{equation} In your case, we have $f(t) = e^{-\lambda t}$ and thus \begin{equation} \int_0^t f^2(u) \mathrm{d}u = \...


8

Of course making money is always the key issue. That (not completely facetious) comment aside: On the practical side, in many firms IT is struggling with being clear, transparent, and intuitive in their handling of multiple curves and their associated risks. Stumbling over your own systems is an annoying way to lose money. These risks can be surprisingly ...


7

The classic argument using risk-neutral pricing is to assume that discounted stock prices are $\tilde{P}$-martingales where $\tilde{P}$ is the risk-neutral probability measure. Then, you know that $$\frac{S_t}{(1+r)^t}=\tilde{E}[\frac{S_T}{(1+r)^T} | \mathcal{F}_t]$$ by definition of a martingale process. As the discounts are non-stochastic, you can ...


7

Multi-fractal models can be applied to the modeling and forecasting of volatility. I read the following book with much interest and actually setup couple models in order to compare performance vs Garch family models and the application of multi-fractals much better captures discontinuous regime-changes than traditional volatility models. Multifractal-...


7

The part where you say that $$\frac{dS_t}{S_t} = d\ln(S_t)$$ is wrong, because $S$ is a stochastic variable. This is exactly what Itô tells you with his formula that you apply right do compute your $dZ$. The difference comes from the quadratic variation of the process $S$ which you express as $(dS)^2$. If you don't add this term when the variable are ...


7

If you consider $X_1$ a random variable which is normally distributed with mean $\mu$ and variance $\sigma^2$ them $S_1 = \exp(X_1)$ is log-normally distributed with mean $\exp(\mu + \sigma^2/2)$ and variance $(\exp(\sigma^2)-1)\exp(2\mu+\sigma^2)$. This follows from the definitions of the normal distribution and the log-normal distribution and deriving the ...


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