# Tag Info

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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