51
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}\\
&= \...
23
These are all examples on Ito Formula in its general form (with quadratic variations):
18
Brownian motion is simply the limit of a scaled (discrete-time) random walk and thus a natural candidate to use. It is very intuitive and arguably one of the simplest and best understood time-continuous stochastic processes. Also, don't forget that you obtain many more stochastic processes as functions of a (time-changed) Brownian motion. In many books on ...
16
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 ...
15
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}\...
14
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 ...
13
Let
\begin{align*}
Y_t = e^{(a+\frac{c^2}{2})t-cW_t}.
\end{align*}
Then
\begin{align*}
dY_t = Y_t\left[\big(a+c^2\big)dt -c dW_t \right].
\end{align*}
Moreover,
\begin{align*}
d(X_tY_t) &= Y_t dX_t + X_t dY_t + d\langle X, Y\rangle_t\\
&=abY_tdt.
\end{align*}
That is,
\begin{align*}
X_t = Y_t^{-1}\left(X_0 + ab\int_0^t Y_sds\right).
\end{align*}
12
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 ...
12
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 ...
12
Assume deterministic and constant interest rates.
For an investor in the foreign economy i.e. a market participant that can only trade assets delivering a payout in the foreign currency, let us define
$$ \tilde{X}_t = \tilde{X}_0 \exp \left(\left(r_f-r_d-\frac{\sigma_\tilde{X}^2}{2}\right)+\sigma_\tilde{X} W_t^{\tilde{X},\mathbb{Q}^f} \right) $$
$$ Y_t =Y_0\...
12
Quadratic variation and variance are two different concepts.
Let $X $ be an Ito process and $t\geq 0$.
Variance of $X_t$ is a deterministic quantity where as quadratic variation at time $t $ that you denoted by $[X,X]_t $ is a random variable.
What is confusing you is the fact that when $X $ is a martingale then
$X^2_t-[X,X]_t$ is a martingale thus you ...
11
I will try to answer this a bit differently.
The rigorous answer: because Ito calculus tells us that we need the second order term. Look at
$$
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:
$$\int_0^t f(t, \omega) d W(t, \omega) = \lim_{\| \Delta_n\| \to 0 } \sum_{i=1}^{n} f(\tau_i,\omega) \left ( W(t_i, \omega) - W(t_{i-1}, \omega) \right ) $$
, ...
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 \...
11
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 = \...
11
Just a few notes
How to make sense of $\text dW_t$ is the entire point of stochastic calculus. It's far beyond the scope of any answer here. You should read some introductory lecture notes/books on stochastic calculus. You could start here.
The idea: Riemann-Stieltjes integrals are of the form $\int_0^t f(s)\mathrm{d}g(s)$ and are well-defined if $f$ is ...
11
By construction, the Itô integral, $I_t=\int_0^t X_s\text{d}W_s$, is a martingale if $\int_0^t \mathbb{E}[X_s^2]\text{d}s<\infty$.
The martingale property, $\mathbb{E}_s[I_t]=I_s$ implies $\mathbb{E}[I_t]=I_0=0$.
Because $W_s\overset{d}{=}\sqrt{s}Z$, where $Z\sim N(0,1)$, we indeed have
\begin{align*}
\int_0^t\mathbb{E}\left[\frac{1}{(1+W_s^2)^2}\right]\...
11
You need to rotate them so we can find some orthogonal axes.
A simple way to think about this is by remembering that we can decompose the second of two brownian motions into a sum of the first brownian and an independent component, using the expression
\begin{align}
W_{t,2} = \rho_{12} W_{t,1} + \sqrt{1-\rho_{12}^2} \tilde{W}_{t,2}
\end{align}
where $\tilde{...
11
Besides @StackG's splendid answer, I would like to offer an answer that is based on the notion that the multivariate Brownian motion is of course multivariate normally distributed, and on its moment generating function.
We know that
$$
\mathbb{E}\left(W_{i,t}W_{j,t}\right)=\rho_{i,j}t
$$
i.e. an $N$-dimensional vector $X$ of correlated Brownian motions has ...
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
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):
$$[Y,X]=[W_t,-\int_0^t\Theta_udW_u]=-\int_0^t\Theta_ud[W_u,W_u]=-\int_0^t\...
10
For any $s \geq t$, note that
\begin{align*}
r_s = r_t + \sigma\int_t^s dW_u + \int_t^s \theta_u du.
\end{align*}
Then,
\begin{align*}
\int_t^T r_s ds &= (T-t)r_t + \sigma\int_t^T\int_t^s dW_u ds + \int_t^T \int_t^s\theta_u du ds\\
&=(T-t)r_t + \sigma\int_t^T\int_u^T ds\, dW_u +\int_t^T\int_u^T\theta_u ds du\\
&=(T-t)r_t + \sigma\int_t^T (T-u)...
10
Just to add to the already nice answers, the result can also be obtained using the (stochastic) Fubini theorem.
\begin{align}
\int_0^t W_s ds &= \int_0^t \int_0^s dW_u\, ds \tag{$W_s=\int_0^s dW_u$}\\
&= \int_0^t \int_u^t ds\,dW_u \tag{Fubini} \\
&= \int_0^t (t-u) dW_u \tag{$\int_u^t ds = t-u $}
\end{align}
And we fall back on the same equation ...
10
Physical objects move according to simple smooth curves that can be represented by low order polynomials: a straight line, a parabola, an ellipse, etc.
Financial market prices move in a completely different way, as can be seen by looking at any graph of stock prices, interest rates etc. in a newspaper: there are constant, erratic fluctuations, sometimes in ...
10
It is, of course, possible to price such a contract in a no-arbitrage market. Indeed, if $f$ is a sufficiently smooth function, then you can price all contracts paying $f(S_T)$. Note that your specific payoff has no optionality and that the payoff may be negative. Bakshi and Madan (2000) discuss the economic meaning of a derivative paying $\cos(S_T)$ in the ...
9
The part where you say that
$$\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 ...
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