55
votes
Integral of Brownian motion w.r.t. time
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. ...
24
votes
Accepted
Worked examples of applying Ito's lemma
These are all examples on Ito Formula in its general form (with quadratic variations):
18
votes
Accepted
Why is Brownian motion useful in finance?
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-...
16
votes
Accepted
Why is this stochastic integral a martingale?
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}...
15
votes
What is the trickiest thing to get right in Rates Quant recently (2019)?
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 ...
13
votes
Accepted
Girsanov Theorem for Quanto/Compo adjustment
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
...
13
votes
Accepted
Solve the following SDE: $\mathrm{d}X_t = a(b-X_t) \,\mathrm{d}t + c X_t \, \mathrm{d}W_t$
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 ...
13
votes
Accepted
Clarification on Deriving Ito's Lemma
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 ...
12
votes
Geometric Brownian motion - Volatility Interpretation (in the drift term)
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, $E[...
12
votes
Why the expected return rate of a stock has nothing to do with its option price?
Because you can hedge. Once you have delta hedged, the pay-off is symmetric about up and down moves so drift doesn't matter.
Also the delta-hedged call and the delta hedged put have to have the same ...
12
votes
Integral of Brownian motion w.r.t. time
Just to add to the already nice answers, the result can also be obtained using the (stochastic) Fubini theorem.
\begin{align}
\int_0^t W_s ds &= \int_0^t \int_0^s dW_u\, ds \tag{$W_s=\int_0^s ...
12
votes
Accepted
Can I always use quadratic variation to calculate variance?
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 ...
11
votes
Accepted
Geometric Brownian motion - Volatility Interpretation (in the drift term)
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 $S_0$ ...
11
votes
How to use the stock as a numeraire to price a derivative with payoff of the form $(S_T f(S_T))^+$?
Let $P$ be the risk-neutral measure. We define the measure $P_S$ such that
\begin{align*}
\frac{dP_S}{dP}\big|_t &=\frac{S_t}{e^{rt}S_0}\\
&=e^{-\frac{1}{2}\sigma^2 t+\sigma W_t}.
\end{align*}
...
11
votes
Why Ito calculus?
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:...
11
votes
Accepted
Ho and lee derivation for short rates model
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 +...
11
votes
Accepted
How were these SDE derived?
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^...
11
votes
Accepted
What is an adapted process
Let $\{X_t\}$ be a stochastic process and $\mathcal{F}$ be a filtration.
The intuitive idea is that for $\{X_t\}$ to be adapted, it can't reveal what's unknowable (according to the filtration). By ...
11
votes
Two papers - two different solutions of the Ornstein-Uhlenbeck process
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, \...
11
votes
Accepted
Expectation of $\int_0^t \frac{1}{1+W_s^2} \text dW_s$
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{...
11
votes
Expectation of exponential of 3 correlated Brownian Motion
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 ...
11
votes
Accepted
Expectation of exponential of 3 correlated Brownian Motion
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 ...
10
votes
Accepted
Girsanov Theorem and Quadratic Variation
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: $$[Y,X]=-\int_0^...
10
votes
Why is this stochastic integral a martingale?
I aim to give a careful mathematical treatment to this answer, whilst following the fantastic book "Basic Stochastic Processes" by Brzezniak and Zastawniak.
The reason I am putting this answer on is ...
10
votes
Accepted
Correlation coeffitiont between two stochastic processes
if you talk about correlation then:
compute expectation:
$$\mathbb{E}(W_t)=0\text{ and }\mathbb{E}(\int_0^tW_d ds)=0$$
variance:
$$\text{Var}(W_t)=t\text{ and }\text{Var}(\int_0^tW_s ds)=\frac{t^3}{...
10
votes
Why is Brownian motion useful in finance?
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 ...
10
votes
Accepted
Periodic functions when determining No Arbitrage price
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 ...
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