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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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I think one has to distinguish between pricing and fitting/reproducing empirical stock returns. A model might fit the empirical stock returns extremly well but fail to reproduce derivative prices. In my answer I will assume that you are interested in reproducing the empirical stock returns. Mandelbrot and the Stable Paretian Hypothesis The most salient ...

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Since both $ER$ and $S$ are gaussian random, why not just assume their dependence is captured by their covariance, and make your draws from the bivariate normal distribution? It is hard to construct any other way of making two marginal gaussians cointegrated. Even if the variables were not gaussian, you would probably find yourself relating them using a ...

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This is simply the integral of the pdf from -0.5 to 0.5 (scaled to the SD of the distribution), also known as the cumulative distribution function or cdf. The cdf(x) function is indicated on the following wikipedia link: Normal Distribution. The normal cdf(x) function computes the integral on [-Infinity, x], so to compute on your interval [x1,x2], is ...

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Starting from the Black-Scholes model that $$\dfrac{dS}{S} = \mu \:dt + \sigma\:dW_t$$ where $W_t$ is a standard Brownian motion, and $\sigma$ and $\mu$ are constant where $\sigma > 0$. Here $W_t$ is a Brownian motion under the physical measure $\mathbb{P}$. We can then use Girsanov's theorem to change the measure to risk neutral measure $\mathbb{Q}$ ...

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Thanks to @Phun and @oliversm I solved the problem. So I'm posting here the solution in case someone will need it. Under Black-Scholes assets dynamics are determined by a Geometric Brownian Motion, and we can define the price of a security at time $t+\Delta t$ as: $$S_{t+\Delta t}=S_{t}\exp\left(\left(r-\frac{1}{2}\sigma^{2}\right)\Delta t+\sigma\sqrt{\... 5 Let (X_t)_{t\geq 0} denote a Geometric Brownian Motion$$ \frac{dX_t}{X_t} = \mu_X dt + \sigma_X dW^X_t,\ \ \ X(0) = X_0$$such that X_t is lognormally distributed \forall t > 0$$ X_t = X_0 e^{(\mu_X - \frac{1}{2}\sigma_X ^2)t + \sigma_X W_t^X}$$Let (Y_t)_{t\geq 0} denote an Arithmetic Brownian Motion$$ dY_t = \mu_Y dt + \sigma_Y dW_t^Y,\ \ \ ...

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So we have the BS-Model $$dS_t=S_t(\mu dt +\sigma dW_t)$$ W.l.o.g we assume $S_0=1$. Itô's lemma implies that $$S_t=\exp{(\sigma W_t+(\mu-\frac{1}{2}\sigma^2)t)}$$ We know that $W_t$ is normally distributed with mean $0$ and variance $t$. Now have a look at the r.v. $$X_t=\sigma W_t+(\mu-\frac{1}{2}\sigma^2)t$$ $\sigma W_t$ is the random part and $\... 4 For small changes, the log-return$\ln \frac{S_{t_i}}{S_{t_{i-1}}}$is close to the simple return$\frac{S_{t_i}-S_{t_{i-1}}}{S_{t_{i-1}}}: \begin{align*} \ln \frac{S_{t_i}}{S_{t_{i-1}}} &= \ln \Big(1+ \frac{S_{t_i}-S_{t_{i-1}}} {S_{t_{i-1}}} \Big)\\ &\approx \frac{S_{t_i}-S_{t_{i-1}}}{S_{t_{i-1}}}. \end{align*} Note also that, assuming the SDE \... 4 My take on the whole issue is as follows: We cannot be sure to find the one and only true model, the only thing we can do is to identify the most prevalent so called stylized facts and try to model them parsimoniously. The following paper was already mentioned in the comments: Empirical properties of asset returns: stylized facts and statistical issues by ... 3 Generally Kurtosis measures the degree to which a distribution is more or less peaked than a normal distribution. Positive kurtosis indicates a relatively peaked distribution. Negative kurtosis indicates a relatively flat distribution. In time series we can encounter high kurtosis which is caused by "fat tails" (higher frequencies of outcomes) at the ... 3 I think there are a few conflating ideas here. With respect to the sum of logs idea, I think you're thinking about infinitely divisible distributions (https://en.wikipedia.org/wiki/Infinite_divisibility_(probability)). These ideas are indeed used to build more complicated models (i.e. Levy processes) for asset returns. With regards to the Efficient ... 3 Perhaps an answer coming from a different angle and giving you some perspective: The typical approach taken by statistics is top-down: Just looking at the data and finding patterns and stylized facts (like excess volatility, volatility clustering, fat tails, no autocorrelation in returns but significant autocorrelation in absolute returns etc.) The problem ... 3 It's not possible with a simple linear transformation like the one you mentioned: since scale and thus the distance between mean and median are required to change, either the mean or the median will not be preserved. Therefore you must use nonlinear transformations, which will complicate quite a bit mantaining skew and kurtosis and imho will not be ... 3 You know that :X \sim N(\mu,\sigma^2)$.$Z = \large\frac{X-\mu}{\sigma}$.$\text{Var}(Z) = \large\frac{1}{\sigma^2}\text{Var}(X) = \large\frac{1}{\sigma^2}\sigma^2 = 1$. So that$Z \sim N(0,1)$. However note that the pdf evaluated for X and Z have different domains. The following figure illustrate it :$X$is plotted in a) and$Z$in b) ... 3 $$C(u,v) = \mathbb{P}\left(X\leq N^{(-1)}(u),\quad \rho X + \sqrt{1-\rho^2}X^\perp \leq N^{(-1)}(v)\right)$$ 2 Another way of seeing it is that the$-\frac12\sigma^2$is just a correction term that comes from Jensen's inequality. You need this when switching from supposedly symmetric returns (normal distribution) to the skewed price process (log-normal distribution). 2 The term 1/2 * sigma-squared arises through the application of Ito's Lemma. Keep in mind that the assumption is of a stock price that follows geometric BM with a constant drift and volatility. If you set up a delta-hedge portfolio and apply Ito calculus you will end up with an adjustment in the distribution by exactly above term. Another way of interpreting ... 2 I think some some terminology got mixed up here. Let$r_t$,$t=1,\ldots,T$be a series of iid excess returns with the estimated mean excess return$\bar{r}= \sum_{t=1}^Tr_t$. Then the Stutzer Index$S$is defined as$ S=\frac{|\bar{r}|}{\bar{r}}\sqrt{2I_p}$with$I_p$being the "Stutzer Information Statistic",$I_p=\max_\theta -\log(\frac{1}{T}\sum_{t=1}^T ...

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Your SDE has no closed-form solution, so you'll have to apply the Euler method to obtain an approximate terminal distribution. Once you have the terminal distributions, any time series you want to validate has a highly multivariate probability density (due to the fact that each day's data comes from a slightly different distribution). You can transform ...

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Basically, what you are asking is: What is the distribution of $$Y = \prod_{i=1}^n X_i$$ where the $X_i$ are i.i.d. and $X_i \sim N(\mu, \sigma^2)$. In general, $Y$ has a very complicated distribution. Check out the discussion in http://math.stackexchange.com/questions/161757/what-is-the-distribution-of-a-random-variable-that-is-the-product-of-the-two-nor?...

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As @Joshua Ulrich points out your distribution gets wider. Approximately, what you do is, you simulate $$Y = X_1 + \dots + X_n$$ and $X_i$ is standard normal. Of yourse the variance increases with $n$ (and standard deviation with $\sqrt{n}$). But: greater variance does not mean heavier tails as suprises. If you want to put this in an easy number (besides ...

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There are many ways answering this, here is one: We assume the asset price at $t=T$, $S_T = S_{T-1} \times (S_T / S_{T-1})$. Assuming continuous compounding, we can write, $S_T = S_{T-1} \times \exp(R_{T-1})$. Working the same way for the previous period, we get $S_{T} = S_{T-2} \times \exp(R_{T-1}+R_T)$. Working all the way back to the initial value of ...

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As the stock price process $S$ follows a geometric Brownian motion, we have that \begin{align*} S_T &= S_0 e^{(\mu-\frac{1}{2}\sigma^2)\, T + \sigma\, W_T}\\ &= S_0 e^{(\mu-\frac{1}{2}\sigma^2)\, T + \sigma\, \sqrt{T}\, \xi}, \end{align*} where $\xi$ is a standard normal random variable. Then, we have the probability \begin{align*} P(S_T > 95) &...

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At what scale do you see kurtosis? Daily data? Single stocks or indices? Let us not look a single stock data, because you always find crazy stocks whose price process breaks all rules. Talking about daily data of indices: they could be thought of the sum of hourly returns or other returns of high frequency (minute returns, milliseconds ...). What are the ...

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The subordinate return process for log returns is normal (or Gaussian). The kurtosis stems from the "activity rate" of events that move asset prices. When we measure in "clock time" we see kurtosis. However, when we measure in "event times" or "business times" the distribution is normal. The "event time" is a subordinator. Substitute "event time" for "...

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Well, log-normality does not allow you to have negative earnings and companies do have negative earnings. I suggest you to download the earnings data and perform a Jarque-Bera test for normality.

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You know that Brownian motion {W(t)} is a stochastic process with the following properties: (Independence of increments) W(t) − W(s) , for t > s , is independent of the past, that is, of W(u) , 0 ≤ u ≤ s, or of $F_s$ , the σ-field generated by W(u), u ≤ s. (Normal increments) W(t) − W(s) has Normal distribution with mean 0 and variance t − s. This implies ...

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Given its price today, the stock price at time T is lognormally distributed, whereas $lnS_T$ is normally distributed, that is $lnS_T$ ~ $N \Bigr(lnS_0 + (\mu- \frac{\sigma^2}{2}T),\sigma^2T \Bigl)$ see for example Hull - Options, Futures, and other Derivatives. Plugging in the numbers you get $lnS_T$ ~ $N(3.981291519,0.16875)$ Then the probability you ...

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In the Black-Scholes framework, we assume the log returns are normally distributed. This is equal to saying the underlying is log-normally distributed. If you look at Geometric Brownian Motion on wikipedia, you'll see this: The above solution S_t (for any value of t) is a **log-normally distributed** random variable The wikipedia is correct.

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