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}\\ &= \...


19

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

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/...


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: $$\int_0^t f(t, \omega) d W(t, \omega) = \lim_{\| \Delta_n\| \to 0 } \sum_{i=1}^{n} f(\tau_i,\omega) \left ( W(t_i, \omega) - W(t_{i-1}, \omega) \right ) $$ , ...


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

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 ...


10

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}\...


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)...


10

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 \...


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

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 ...


9

Quadratic variation and variance are two different concepts. Let $X $ be an Ito process and $t\geq 0$. Variance of $X_t$ is a deterministic quantity where as quadratic variation at time $t $ that you denoted by $[X,X]_t $ is a random variable. What is confusing you is the fact that when $X $ is a martingale then $X^2_t-[X,X]_t$ is a martingale thus you ...


9

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 ...


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 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 ...


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

Stochastics are usually applied in the field of derivatives pricing. In this setting the task is to price a derivative such that it fits into the landscape of tradable instruments (no-arbitrage). We work using the risk-neutral measure - usually denoted by $Q$. The measure is derived from other traded instruments. In risk analysis (e.g. calculate the VaR, ES ...


8

Here is a solution without using the PDE technique, which is preferred as we do not need to assume the affine form of a zero-coupon price from the start. we assume that, under the risk-neutral measure, \begin{align*} dr_t = (\theta(t)-a r_t) dt + \sigma dW_t, \end{align*} where $a$ and $\sigma$ are constants, $a(t)$ is a deterministic function, and $W_t$ is ...


8

The formula $F^X(t,T) = E_t^d\left(X_T \right)$, under the domestic risk-neutral measure, is problematic. Note that, at time $t$, the forward exchange rate $F^X(t,T)$, for maturity $T$, is the exchange rate such that the payoff $X_T-F^X(t,T)$ has a zero value at $t$. That is, \begin{align*} B_t^d E_d\left(\frac{X_T-F^X(t,T)}{B_T^d} \mid \mathcal{F}_t\right)=...


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

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 = \...


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 ...


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