# Tag Info

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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 ...

15

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): $d\,X_t=\mu X_t dt + ... 10 Baxter and Rennie say it better than me, so I will summarize them. Suppose that$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 ... 10 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 ... 10 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 ... 9 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 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 ... 7 It's a lemma! Ito's Lemma gives the change of coordinates rule for stochastic calculus. The multiplication rule is a shorthand way of expressing it. 7$X_t$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) $$df(t,X_t) = \frac{... 6 I like Richard's answer, but I think we can compute the mean and the variance of \int_0^T W_t dt by ourselves using Ito's lemma. Let f(W_t, t) = t W_t.$$ 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) + \frac{\sigma}{... 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 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 ... 6 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 ... 6 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 ... 6 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 ... 6 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 + ...

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Here's my favorite example of an intraday strategy on S&P500 futures that at least used to work: Intraday Share Price Volatility and Leveraged ETF Rebalancing I pull it out whenever people start talking about market efficiency. The strategy is very simple: if S&P500 futures are up or down more than 2% on the day with two hours left until close, ...

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You seem to use the term "volatility" to describe two very different quantities: (1) the diffusion coefficient of your SDE and (2) the standard deviation of the log-returns under your modelling assumptions. While the first may be negative, the second may not. [Interpretation 1] Consider a probability space $(\Omega,\mathcal{F},\mathbb{P})$ and a standard ...

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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 $S$. ...

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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 (...

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

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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: $$E[d(S_t/... 5$$ \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) $$using Ito's lemma. Then we can sub in the original processes yields \... 5 Brownian motion - because it is simple, and results in intuitive closed form solutions, and it's not a terrible description of asset prices, especially when employed in high-frequency event time. Geometric - because the returns compound, and equities cannot go below zero due to the fact that they are limited liability corporations There are many, many ... 5 For a Brownian motion, if you wait dt, the variance will grow linearly with (proportionally to) dt. For a fractional Brownian motion, it will grow with a power law of dt, in fact in dt^{H}, where H is the Hurst exponent. See wikipedia for more details. It means the fBM will somehow keep memory of the past. When H is lower than 1/2, it will mean ... 5 You ask 2 questions and I try to answer: 1) Why do we use geometric Brownian motion (\ln S_t-\ln S_0 is normally distributed)? In this case you have$$ S_t = S_0 \exp( (\mu-\sigma^2/2) t + \sigma B_t), $$which means that you model positive prices. Furthermore the log-return$$ \ln(S_t/S_0) = (\mu-\sigma^2/2) t + \sigma B_t,  is normally distributed. ...

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The first process is a BM. The second does not exist in continuous time. The variance goes down too slowly with dt and the process blows up at the limit. You can break the (0,1) interval into 1, 100, 1000, 1000000 steps and see that happening. Variance of a martingale has to scale with dt: if it is too fast then the process dies, if it is too slow then ...

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What can be shown is that the above expressions are equal in probability. First check the distribution. As any linear combination of a Gaussian is Gaussian the right hand side is Gaussian - the left hand side too. Then we need the 2 moments: The expected values - it is zero ... easy to see. Next what you did not specify is that the correlation between \$...

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