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## Hot answers tagged geometric-brownian

14

Intuition Yes it is possible. Both, the NIG and the VG process are exponential Lévy processes, i.e. they model the stock price via $S_t=S_0e^{X_t}$, where $X_t$ is a Lévy process. Here's a recent answer to the topic. Your question boils down to the following: if $X_t$ is a general Lévy process (VG, NIG, etc.), can we find parameters of $X_t$ such that $X_t$ ...

9

I provide a solution in three steps. The first step carefully outlines how to split up the expectation and what new measures are used. This first step does not require any special model assumption and holds in a very general framework. I derive a formula for the option price that resembles the standard Black-Scholes formula. In a second step, I assume that ...

5

I think the easiest way to derive the solution to the GBM is via Ito's Lemma. The GBM: $dS_t = \mu S_t dt + \sigma S_t dW_t$ is a short hand for: $$S_t = S_0 + \int_{h=0}^{h=t}\left(\mu S_h\right)dh + \int_{h=0}^{h=t}\left(\sigma S_h\right)dW_h$$ Ito process is defined as: $$X_t = S_0 + \int_{h=0}^{h=t}\left(a(X_h,h)\right)dh + \int_{h=0}^{h=t}\left(b(... 5 You're right but a GBM doesn't assume that percentage returns are normally distributed. It's about log-returns. If the log-return r_t=\ln\left(\frac{S_{t+dt}}{S_t}\right) is normally distributed (GBM assumption), then r_t can indeed be any arbitrarily large (positive or negative) number with positive probability. This also implies that stock prices are ... 5 As shown in Credit Risk Modeling Notes (Bielecki, Jeanblanc, Rutkowski), Corollary 1.3.1, for t < s, we have:$$ P(\tau \leq s | {\cal F}_t) = N\left( -Y_t \sigma^{-1}(s-t)^{-1/2}- \nu(s-t)^{1/2}\right ) + {\rm e}^{-2\nu \sigma^{-2}Y_t} N\left( -Y_t \sigma^{-1}(s-t)^{-1/2}+ \nu(s-t)^{1/2}\right ),$$where$$ Y_t = y_0+ \nu t +\sigma W_t, \: \sigma >0,...

4

It's just Girsanov's theorem. I suppose that under the risk neutral measure Q $$dS_{t}= r S_{t} dt + \sigma S_{t}dW_{t},$$ $$S_{t} = S_{0}\exp\left((r-\frac{\sigma^{2}}{2})T + \sigma W_{T}\right)$$ By multiplying by $e^{-rT}$ I have $e^{-rT}S_{T}$ which is a martingale so that I can change my measure under $Q$ to some equivalent probabilty $Q_{1}$ under ...

4

$\frac{1}{S_t}$ is log-normal If $S_t$ is a geometric Brownian motion, so is $\frac{1}{S_t}$ and indeed any power $S_t^\alpha$. Simply use Itô's Lemma and set $f(t,x)=\frac{1}{x}$, \begin{align*} \mathrm{d}f(t,S_t) &= \left(0-\mu S_t\frac{1}{S_t^2}+\frac{1}{2}\sigma^2S_t^2\frac{2}{S_t^3}\right)\mathrm{d}t-\sigma S_t \frac{1}{S_t^2}\mathrm{d}W_t \\ &=-...

3

Black scholes formula based on $S_t$ measure , theory, and formulas you mention are derived in detail in "Steven Shreve: Stochastic Calculus and Finance" draft pdf from 1997 , page 328 "stock price as numeraire".

3

Heavier tails, or a higher probability of extreme outlier values, meaning the investor is more likely to experience extreme events (e.g. tail losses). EDIT: @noob2, valid point, see this article on modelling and forecasting the kurtosis and returns distribution of financial markets: irrational fractional Brownian motion model approach

3

Following this answer, let $\mathbb Q$ be the probability measure associated to the risk-free bank account as numeraire and $\mathbb Q^1$ the probability measure associated to the stock as numeraire. You know that the standard equation $\mathrm{d}S_t=rS_t\mathrm{d}t+\sigma S_t\mathrm{d}W_t^\mathbb{Q}$ can be written as $\mathrm{d}S_t=(r+\sigma^2)S_t\mathrm{d}... 3 Part 1: deriving the drift of the stock price process under the stock Numeraire. Under the risk-neutral measure, the process for$S_t$is as follows: $$S_t = S_0 + \int_{h=t_0}^{h=t}rS_h dh + \int_{h=t_0}^{h=t}\sigma S_h dW_h = \\ = S_0exp\left[ (r-0.5 \sigma^2)t+\sigma W(t) \right]$$ In the above model, the Numeraire is$N(t)=e^{rt}$with$N(t_0):=1$. ... 3 Question 1 is answered in parts 1 through to 6: the idea is that each part slowly builds the tools required to derive the process equation for$S_t$under the$S_t$Numeraire. Question 2 & Question 3 are then answered in part 7. Part 1: Expectation of a function of a Random variable: Let$X(t)$be some generic Random Variable with probability density ... 3 Echoing some of the comments to the OP above, the only real difference between random walks and Brownian motions is a question of time frequency. IE a Brownian motion is just an aggregation of a (binary) random walk with higher frequency. Given both will always be at best an approximation of reality, asking for which is "better" becomes a bit of a ... 3 The return$R_i$as expressed in $$R_{i+1,i}=\frac{S_{i+1}-S_i}{S_i}=\mu \Delta t + \sigma \Delta W(t_{i+1},t_i)$$ is not possible. To see this, let's get the returns over two small time steps of$\Delta teach. Then $$R_{i+2,i+1}=\frac{S_{i+2}-S_{i+1}}{S_{i+1}}= \mu \Delta t + \sigma \Delta W(t_{i+2},t_{i+1})$$ but $$R_{i+2,i}=\frac{S_{i+2}-S_{i}}{S_{i}}= ... 3 I assume you want to real-world probability, because the risk-neutral probability is not a probability in the 'likelihood' sense. Under the real-world measure, we model the stock under the B-S model as:$$X(t)=X(0)+\int^{t}_{0}\mu X(h)dh+\int^{t}_{0}\sigma X(h)dW(h)$$If the market demands a 30% annual return, I will take that as the real-world rate \mu.... 3 Risk-neutral pricing is to help with relative value type questions: If I know the value of this what should the value of that be if it depends in some way on this. It doesn't help with absolute value type questions: Should I buy this or that, is the implied volatility too low or high etc. Those are generally "real world measure" type questions. 3 Great question! Abstract: Your code and math are correct, but you use too high vol and drift to be real world realistic. Your simulations decay to zero due to high vol and LogNormality. Basically, I used two slightly different approaches. Based on my research, it should be either possible to use daily drift and volatility with dt = 1 or annualize drift and ... 2 No need to use ito's lemma. You can simulate your process directly from the equation:$$dS_t = \mu S_t dt + \sigma S_t^{\beta/2} dW_t$$which means that:$$S_{t_{i+1}}=S_{t_i}+\mu S_{t_i}(t_{i+1}-t_i)+\sigma S_{t_i}^{\beta/2}\sqrt{t_{i+1}-t_i}Z_{i}$$where Z_i is a realization of normal distribution with mean 0 and variance equal to 1. 2 Let n be the number of stocks (here n=3) Let T be the number of sequential returns to generate (for example T=12 if you want to generate a year's worth of monthly returns) Let M be the number of alternative scenarios to generate (for example M=1000 to generate 1000 different outcomes) Then, Step 1. You generate a n \times T matrix RETS of ... 2 Here is a simple solution using the equivalence of no arbitrage and the existence of a stochastic discount factor. Let the SDF be \Lambda(t). This evolves as$$\frac{d\Lambda(t)}{\Lambda(t)}=-rdt-\varphi(t) dW(t),$$where we used the fact that the drift of the SDF is the risk-free rate and that there is only one source of uncertainty. The standard pricing ... 2 Call option:$$\mathbb{P}\left(S_t\geq K\right)=\mathbb{P}\left(S_0e^{(rt-0.5\sigma^2t+\sigma W_t)}\geq K\right)=\\=\mathbb{P}\left(W_t\geq \frac{ln\left(\frac{K}{S_0}\right)-rt+0.5\sigma^2t}{\sigma}\right)=\\=\mathbb{P}\left(Z\geq \frac{ln\left(\frac{K}{S_0}\right)-rt+0.5\sigma^2t}{\sigma\sqrt{t}}\right)=\mathbb{P}(Z\leq d2)$$So we have shown the well-... 2 We know the formula to price a call option in the Black-Scholes-Merton model:$$C=S_0\Phi(d_1)-e^{rt}K\Phi(d_2)$$with d_1=\frac{\log\frac{S_0}{K}-T(r+\frac{\sigma^2}{2})}{\sigma\sqrt T} and d_2=d_1-\sigma\sqrt T, assuming the underlying stock pays no dividends. The option delta is given by:$$\Delta:=\frac{\partial C}{\partial S}=\Phi(d_1)$$Note that ... 2 For simplicity, we suppose r = 0.$$V_t = E^Q((S_T^1 - K)1_{\{S_T^1 > K\}}1_{\{L<S_T^2<U\}}) = E^Q(S_T^11_{\{S_T^1 > K\}}1_{\{L<S_T^2<U\}}) - KE^Q(1_{\{S_T^1 > K\}}1_{\{L<S_T^2<U\}})$$For the second term:$$E^Q(1_{\{S_T^1 > K\}}1_{\{L<S_T^2<U\}})=P(\{S_T^1 > K\}\cap \{ L<S_T^2<U \}) =P(\{(W_T^1-W_t^1) > d_1 \... 1 As per the hint, you first write \begin{align*} \mathbb{E} \left [\left (\int_{0}^{t} W_{s}^{2}\, ds \right )^{2} \right ] &= \mathbb{E} \left [\left (\int_{0}^{t} W_{s}^{2}\, ds \right )\left (\int_{0}^{t} W_{u}^{2}\, du \right ) \right ] \\ &= \mathbb{E} \left [\int_{0}^{t} \int_{0}^{t} W_{s}^{2}W_{u}^{2}\, du \, ds \right ] \\ &= \mathbb{E} \... 1 Another sketch of proof: If you move to the equivalent PDE (using Feynman-Kac), you can assume that S is positive, find the solution by log-transfomation. Then as the solution is unique given initial conditions, and it is the solution of the original PDE, S must be positive. 1 You may want to read this paper for GBM with discrete dividend. 1 The shifted exponential of a lognormal distribution, just as the exponential of a lognormal distribution is a known in finance because of zero-coupon bond option. I am not aware of it being otherwise known. You can use a lognormal assumption on the bond price instead of the yield. This will allow you to use the Black formula. You can approximately fit the ... 1 Try the code below. It consumes less memory and takes about 6 minutes to run. Please note that when simulating the stock price for purposes of pricing derivatives, we use the risk-neutral stock price process. The drift of the risk-neutral process becomes the risk free rate minus half the square of annualized volatility as indicated in the code. import numpy ... 1 Echoing the comments earlier in the thread, I suggest we work backwards by first picking the key aspect of the market (e.g. flash crashes, liquidity shocks, long-memory) we seek to replicate. Then we select the stochastic model that best approximates our target behavior. We'll need to decide how to calibrate our models to the market and picking a salient ... 1 You can do what we always do and take logs and Itô's Lemma: $$\text{d}\ln(X_t)= \left( b(t)-\frac{1}{2}\sigma^2(t)\right)\text{d}t+\sigma(t)\text{d}B_t.$$ Then, by definition, $$\ln(X_t)=\ln(X_0)+\int_0^t\left( b(s)-\frac{1}{2}\sigma^2(s)\right)\text{d}s +\int_0^t \sigma(s)\text{d}B_s$$ or\$X_t=X_0\exp\left(\int_0^t\left( b(s)-\frac{1}{2}\sigma^2(s)\right)\...

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