Tag Info

Hot answers tagged

11

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

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 ...


9

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 ...


8

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 ...


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 ...


6

Martingales + Markovian Here is the motivation. Conditional expectations are martingales by the tower property of conditional expectations (an easy exercise to show). Suppose $r=0$, by the risk neutral pricing theorem $E^\star\left[h(X_T)\bigg|\mathscr{F}_t,\,X_t=x\right]$ is the price of any derivative security with $X$ as the underlying asset and payoff ...


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 ...


5

The Feynman-Kac theorem primarily makes sense in a pricing context. If you know that some function solves the Feynman-Kac equation you can represent it's soluation as an Expectation with respect to the process. (confer this document) On the other hand a pricing function solves the FK-PDE. Thus often one would try solving the PDE to get a closed form ...


5

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 ...


5

$$ \textbf{Preface} $$ I am assuming log normal asset but this is not clear from the question? Or rather I have misinterpreted the question! Well as I see it from a a purely mathematical exercise $$ d\left(\dfrac{S_t}{M_t}\right) =\frac{1}{M_t}dS_t - \frac{S_t}{M_t^2}dM_t +O(dt^2) $$ using Ito's lemma. Then we can sub in the original processes yields ...


5

The trick is to start with the highest power, rewrite it as something you know (a third order moment) and then work backwards on the remaining terms. By that I mean you can complete the cube as follows: $$E[W_t^3 - 3tW_t|\mathcal{F}_s] = E[(W_t-W_s)^3 - C -3tW_t|\mathcal{F}_s]$$ where you'll need to find $C$ such that the equality holds (i.e. $C=W_s^3 + ...


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.


4

I would calculate it this way, $\mathbb{E}[(W_s+W_t−2W_0)^2] = \mathbb{E}\left[\left((W_s-W_0)+(W_t-W_0)\right)^2\right]\\ \hspace{4cm}=\mathbb{E}[(W_s-W_0)^2]+\mathbb{E}[(W_t-W_0)^2]+2\mathbb{E}[(W_s-W_0)(W_t-W_0)] \\ \hspace{4cm}=s+t+2\mathbb{E}[W_sW_t]\\ \hspace{4cm}=s+t+2\min(s,t)$


4

The formula is given in your link. For the real world probability without jump: $$x_t = x_{t-1} e^{-\eta \Delta t} + \hat{x}(1-e^{-\eta \Delta t}) +\sigma \sqrt{\frac{1-e^{- 2 \eta \Delta t}}{2 \eta}} N(0,1) $$ where: $x_t$: price $x_{t-1}$: PreviousPrice $\hat{x}$: long term mean (a parameter) $\Delta t$: Time step (one fraction) $\eta$: ...


4

Most of the time, when you have a simple SDE without a drift, it's a martingale because the Wiener process itself is a martingale. In your example, you have a constant with the Wiener process, therefore the whole process must also be a martingale because the expectation is clearly X(t). However, we can't conclude a driftless SDE is always a martingale. ...


4

If at first you don't have a model at all, then geometric Brownian motion is not bad. As others before me said: log-returns are normally distributed in this model. This is debatable and there are times and markets where this is not true. There is more than enough research about this. But why is a model based on Brownian motion not that bad? The reason is ...


3

Maybe this could also be a comment but I think an it is not possible to answer this question with a 'yes and here is how you do it'. It has been tried, e.g. by me for a university research project. In this research we focused primarily on aggregation of returns and the main problem was the tractability of the resulting distributions and expressions, also ...


3

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 ...


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

It is a binomial tree in disguise.


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 ...


3

Question 2 has a straight forward solution using a differential equation approach: $\mathbb{P}(\tau^\mu_a<\infty)=1$ The following link (pp. 21 f.) explains it nicely (and is also very detailed) - could not write it much better. If you were to google "brownian motion linear boundary" you will get additional results. Also if you are generally interested ...


3

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: $$ ...


3

A key property of Brownian motion is independent increments. So if $x-1 > y$, then $$ \mathbb{E}[\Delta W_x \Delta W_y] = 0 $$ because the time intervals [x-1,x] and [y-1,y] do not overlap. If they do overlap, i.e. $x-1 \leq y < x$, then \begin{align} \mathbb{E}[\Delta W_x \Delta W_y] =&\ \mathbb{E}[(W_x - W_{x-1}) (W_y-W_{y-1})] \\ =&\ ...


3

The Papageorgiou paper is presumably referring specifically to quasi-random sequences used in path generation. Researchers had noticed that, in high dimensions, QR sequences tend to have good space coverage for the first couple of dimensions: but terrible coverage for the latter dimensions: (Plots here are points 101-200 from a 32-dimensional QR ...


3

To solve the expectation directly, you need to remember that a density function is not the same as the probability of the event. We have, $\frac{S_1}{S_0} \sim \ln \mathcal{N} \left(-\frac{\sigma^2}{2},\sigma\right)$, therefore, \begin{eqnarray} \mathbb{E}\left(\frac{S_1}{S_0}\right) &=& \int_{-\infty}^\infty x\, f_{\frac{S_1}{S_0}}(x)dx\\ ...



Only top voted, non community-wiki answers of a minimum length are eligible