28
The way you do it in the first place is a discretization of the Geometric Brownian Motion (GBM) process. This method is most useful when you want to compute the path between $S_0$ and $S_t$, i.e. you want to know all the intermediary points $S_i$ for $0 \leq i \leq t$.
The second equation is a closed form solution for the GBM given $S_0$. A simple ...
18
Here is a short list (to be edited and improved - community wiki) :
Standard brownian motion (also called Wiener process) for which:
$d\, W_t \sim \mathcal N(0, \sqrt{d t})$
Geometric brownian motion, used in the Black-Scholes model (1973):
$d\,X_t = \mu X_t\,dt + \sigma X_t\,dW_t$
Constant elasticity of variance ("CEV") model (1975):
$d\,X_t=\mu X_t dt + \...
17
Using the Ito Formula
The general approach that often works for these kinds of question is to search for functions such that their Ito differential contains the terms that we are interested in. In your case, we are looking for a function $f(t, x)$ such that $f_t(t, x) = t x$. Let
\begin{equation}
f(t, x) = \frac{1}{2} t^2 x
\end{equation}
with
\begin{...
14
Basically, prices usually have a unit root, while returns can be assumed to be stationary. This is also called order of integration, a unit root means integrated of order 1, I(1), while stationary is order 0, I(0). Time series that are stationary have a lot of convenient properties for analysis. When a time series is non-stationary, then that means the ...
12
To complement @SRKX comment ,i'll try to explain the "simple mathematical proof" beetween both formula :
I assume you know the geometric or arithmetic brownian motion :
Geometric:
\begin{equation*}
dS = \mu S dt + \sigma Sdz
\end{equation*}
Arithmetic :
\begin{equation*}
dS = \mu dt + \sigma dz
\end{equation*}
Then another important stochastic tool you ...
12
I think to understand the martingale/local martingale distinction, it helps to bring in a third class of processes, the uniformly integrable martingale. I would argue that the local martingale and the non-uniformly integrable (true) martingale are actually fairly similar.
The key property that a uniformly integrable martingale has is the so-called closure ...
11
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 ...
11
I will assume a white noise is a process $(\varepsilon_t)$ with zero mean, no autocorrelation and constant variance $\sigma^2 > 0$ while a random walk is a process $(x_t)$ defined by
$$
x_{t+1} = x_t + \varepsilon_{t+1}
$$
where $\varepsilon$ is a white noise.
1) No since $Var(x_{t+1}) = Var(x_t) + Var(\varepsilon_{t+1})$ is stricly increasing while ...
11
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
Another approach consists in using the Fubini theorem to write that
\begin{align}
\int_0^T u W_u du &= \int_0^T \int_0^u u\, dW_v\, du \tag{$W_u = \int_0^u dW_v$} \\
&= \int_0^T \int_v^T u\, du\, dW_v \tag{Fubini}\\
&= \frac{1}{2}\int_0^T (T^2 - v^2) dW_v
\end{align}
This is an Itô integral. Since the integrand ...
11
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*}
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:
$$
E[d(S_t/...
10
Because you can hedge. Once you have delta hedged, the pay-off is symmetric about up and down moves so drift doesn't matter.
Also the delta-hedged call and the delta hedged put have to have the same value since they have the same pay-off. (Put-call parity) Yet any argument that the call should be worth more because of drift says that the put should be ...
10
If you want to address interesting problems that are interesting for financial mathematics, I do not believe you have the good list.
Pricing.
For instance, most of explicit formulas for pricing that are not available yet will never be. In this direction, you should have a look at simulation techniques. See for instance Nonlinear Option Pricing. Interesting ...
9
This is an interesting question that I have asked myself. Below is my take.
Let us consider an economy $(\Omega,\mathcal{F},P)$ equipped with a filtration $(\mathcal{F})_{t \geq 0}$ consisting on a traded asset $S_t$ and a numéraire $N_t$ specified by the following stochastic differential equations:
$$\begin{align}
\text{d}S_t&=\alpha(t,S_t)\text{d}t+\...
9
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 = \...
9
I will provide some references such that you can see where the different processes are used. These papers typically motivate their models and show which effect the single paramaters have and what asset price dynamics the model intends to capture.
Geometric Brownian motion
The geometric Brownian motion is the easiest model for exponential growth with ...
9
I've seen that Gordon answer is more concise and to the point. Take this as a complementary answer.
This is a general approach that will work for all this type of linear SDEs, not just this one. Assume we have the following linear SDE
$$dX_t = (F_t X_t +f_t)dt + (G_t X_t +g_t)dB_t \tag*{(1)}$$
where $F, G, f$ and $g$ are Borel measurable bounded ...
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:
$$ S_t = S_0 \...
8
We know that $(\tilde{W}_t) := (-W_t)$ is also a Wiener process so
$$
E[W_pW_qW_r] = E[\tilde{W}_p\tilde{W}_q\tilde{W}_r] = (-1)^3E[W_pW_qW_r]
$$
and that implies that $E[W_pW_qW_r] = 0$.
8
Stochastics are usually applied in the field of derivatives pricing. In this setting the task is to price a derivative such that it fits into the landscape of tradable instruments (no-arbitrage). We work using the risk-neutral measure - usually denoted by $Q$. The measure is derived from other traded instruments.
In risk analysis (e.g. calculate the VaR, ES ...
8
$X_t$ being a stochastic process, one cannot use ordinary calculus to express the differential of a (sufficiently well-behaved) function $f$ of $t$ and $X_t$.
Instead one should turn to Itô's lemma, one of the key results of stochastic calculus, which stipulates (assuming $X_t$ is here a continuous, square integrable stochastic process)
$$ df(t,X_t) = \frac{...
8
As Sanjay said, you can apply Itô's Lemma to $f(t,x)=x^2$ and obtain
\begin{align*}
\mathrm{d} S^2_t=\left(2\mu S_t^2+\sigma^2S_t^2\right)\mathrm{d}t+\left(2\sigma S_t^2\right)\mathrm{d}W_t.
\end{align*}
Thus, $(S_t^2)$ is again a geometric Brownian motion and hence, for each time point $t$ log-normally distributed with drift $2\mu+\sigma^2$ and volatility $...
7
To solve this equation, let
\begin{align*}
M_t = e^{(\theta + \frac{1}{2}\sigma^2 ) t - \sigma W_t}.
\end{align*}
Then
\begin{align*}
dM_t = M_t\Big[\big(\theta +\sigma^2\big) dt - \sigma dW_t\Big].
\end{align*}
Moreover,
\begin{align*}
d(M_t X_t) &= M_t dX_t + X_t dM_t + d\langle M, X \rangle_t\\
&=\theta\,\mu\, M_t dt.
\end{align*}
Then,
\begin{...
7
There is a lot of ways to understand why stationarity allows to apply usual time series analysis. Here is one more.
Very often, the theoretical justification of what you do in time series need to be able to identify the mean formula and the expectation:
$$\frac{1}{N}\sum_{n=1}^N X_n \underset{N\rightarrow +\infty}{\longrightarrow} \mathbb{E} X, $$
where the ...
7
To provide a straight forward answer: It is not a good model. It never was, it never will be. Until we all do not come up with a better model that provides better modeling accuracy while it is equally intuitive and makes similarly simplifying assumptions the BS model with its geometric brownian motion component is here to stay.
It actually does not matter ...
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