Hot answers tagged brownian-motion
9
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 ...
8
One reference is "The Econometrics of Financial Markets" by John Y. Campbell, Andrew W. Lo, & A. Craig MacKinlay -- http://press.princeton.edu/TOCs/c5904.html. In particular:
9.3.1 Parameter Estimation of Asset Price Dynamics 356
9.3.4 The Effects of Asset Return Predictability 369
You might also take a look at Chan (1992) "An Empirical Comparison ...
7
Yes, you need Cholesky factorization.
You can find the general idea here:
http://www.goddardconsulting.ca/option-pricing-monte-carlo-basket.html
Plus the implementation in MATLAB here:
http://www.goddardconsulting.ca/matlab-monte-carlo-assetpaths-corr.html
The code in general should be easily translatable. The only difficulty is the Cholesky factorization ...
7
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 ...
6
I like Richard's answer, but I think we can compute the mean and the variance of $\int_0^T W_t dt$ by ourselves using Ito's lemma. Let $f(W_t, t) = t W_t$.
$$
d( t W_t ) = W_t dt + t dW_t .
$$
Integrating both sides, and re-arranging the terms, we get
$$
\int_0^T W_t dt = T W_T - \int_0^T t dW_t \, .
$$
We'll be using Ito's isometry formula $\mathbb{E} ...
6
The model for the stock is the Bachelier model with the solution
$$
S(t) = S(0) + \sigma W(t)
$$
Thus the law of the stock $S(t)$ is Gaussian with mean $S(0)$ and variance $\sigma^2 t$.
For average process $Z(T)$ is thus the average of linear Brownian motion, we can rewrite this as
$$
Z(T) = \frac{1}{T} \int_0^T S(0) + \sigma W(t) dt = S(0) + ...
6
well, it is absolutely in agreement with theory. the correlation as measured by Pearson's coefficient $\rho$ is linear measure in the sense that the bounds [-1,1] are obtained only when transformations of our variables are linear, so if we have variables $X$ and $Y$ then something like $aX+bY+c$ where $a,b\in\mathbb{R^*}$, $c\in\mathbb{R}$ will have ...
5
For completeness, let's restate that the discrete case goes like this:
$$\Delta S_t = S_{t+\Delta t}- S_t = \mu S_t \Delta t + \sigma \sqrt{\Delta t} Z_t $$
with $Z_t \sim \mathcal{N}(0,1)$.
What you are doing in your case (although there is a typo in your formula) is to use the exact solution of the SDE to model the move between two points of $S$.
...
5
As you said, $\mu$ is the expected return that is the expected value (mathematical expectation) of the random variable "stock return" under the objective probability measure. Assuming that returns are stationary*, the obvious way to estimate it is to compute a large number $N$ of returns $R_i$, then to average them. You also want to annualize this average ...
4
So where to begin? Continuity is a big thing as it fails to take into account jumps, the Gaussian assumption is another big one. However, looking deeper into it stationarity is a huge problem as it applies to financial time series.
However, it does an OK job at simulation stuff in the long-run.
3
Note: There is a typo in your third equations. Instead of $S(u)$ it should be $S(t_{i})$ and in place of $S(t)$ there should be $S(t_{i+1})$.
In fact, given $S(t_{i})$ we have that
$$S(t_{i+1}) = S(t_{i}) \exp\left( (\mu - \frac{1}{2} \sigma^2) (t_{i+1} - t_{i}) + \sigma (W(t_{i+1}) - W(t_{i})) \right)$$
is the exact solution of the SDE. Hence, the ...
3
From this paper:
The geometric Brownian motion model implies that the series of first differences of the log prices must be uncorrelated. But for the S&P 500 as a whole, observed over several decades, daily from 1 July 1962 to 29 Dec 1995, there are in fact small but statistically significant correlations in the differences of the logs at short time ...
2
I take it that μ is the drift of the long-term equilibrium price.
let's take a lognormal model as an example,
dS = μ x dt + σ x S x dz
where:
S= spot,
t = time,
T-t = length of time
μ = drift rate,
σ = volatility,
dz= random variable,
In order to solve for μ, you might first want to look for the expected spot price:
given X = ln(S),
dX = ((μ - ...
2
It depends on the frequency and the horizon. For instance, I got a similar looking chart when I used annual log returns as the input to the log normal distribution and went out 250 years. With daily log returns over a few years, there isn't nearly as much of a decay. However, when you go out 250 years with daily returns you still see the pattern.
2
I think there is no mistake on your part, if you set sigma <- 0.0045 and
x <- seq(100, 112, length=100) // Lower values produce jagged edges
y <- seq(0.25, 1.1, length=60)
you'll get this:
With these parameters the density has about the same peak and the maximum of the density function also has a similar direction. Alas, a number of things are ...
2
You want to work directly with $\overline{X}$, and not some other r.v. with the same distribution, since equivalence in distribution doesn't imply that correlation remains the same. For ease of notation, I'll assume that $\mu = 0$ and $\sigma = 1$. I claim that
$$
\text{cov}\left(\overline{X},X \right) = \frac{1}{t} \int_0^t s \ ds.
$$
Note that this is ...
2
There is of course no closed-form formula for this. However, the community has long since worked out what all the distributional moments are. A common use is to get the equivalent lognormal (or sometimes shifted lognormal) distribution to a portfolio (such as your difference).
Here's a recent moments paper in which they go so far as to run a binomial tree ...
1
$$
\frac{dS_{1t}}{S_{1t}}=\mu_1 dt + \sigma_1 dW_{1t} \to S_{1t} = S_{1,t=0}e^{\int^t_0 \mu_1 - .5\sigma^2_1 ds + \int^t_0 \sigma_1dW_{1s}}\\
\frac{dS_{2t}}{S_{2t}}=\mu_2 dt + \sigma_1 \rho dW_{1t} + \sigma_2 (1-\rho)dW_{2t} \to S_{2t} = S_{2,t=0}e^{\int^t_0 \mu_2 - .5 (1-\rho)^2\sigma_2^2 + \rho^2 \sigma_1^2 ds + \rho \int^t_0 \sigma_1dW_{1s} + ...
1
Its very simple,
One of Brownian Motion (a.k.a. Wiener process in Mathematics) properties is that each increment from s->t is normally distributed with mean = 0 and sd = t-s.
So, if the process that drives your simulated results is ~N(0, t-s) distributed for each increment s->t with 0<=s<=t then yes, your simulated draw downs should match the ones ...
1
Take a look at the following paper about the Maximum Drawdown distribution:
On the Maximum Drawdown of a Brownian Motion
The authors end up with an approximative series for the density. It is implemented in the function maxdd of the R-package fBasics. There are convenient functions dmaxdd, pmaxdd and rmaxdd. Calculating the Expected Drawdown should be ...
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