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 ...


8

As Sanjay said, you can apply Itô's Lemma to $f(t,x)=x^2$ and obtain \begin{align*} \mathrm{d} S^2_t=\left(2\mu S_t^2+\sigma^2S_t^2\right)\mathrm{d}t+\left(2\sigma S_t^2\right)\mathrm{d}W_t. \end{align*} Thus, $(S_t^2)$ is again a geometric Brownian motion and hence, for each time point $t$ log-normally distributed with drift $2\mu+\sigma^2$ and volatility $...


5

The issue is that you do not plot one sample path but for each time point $t$, you simply plot one possible realisation of the random variable $S_t(\omega)$. Thus, you don't get a connected path. (Just as a minor, you would need brackets in the exponential in your for loop, i.e. X_analytic[i] <- X_analytic[1]*exp((mu - 0.5*sigma^2)*time[i] + sigma*Z[i-1]...


5

In Hull's textbook, the stock price dynamics is lognormal: $S_T = S_0 \exp(\mu T - \frac{1}{2}\sigma^2T + \sigma W_T)$, where $W_t$ is a standard brownian motion. And so the mean of this is the mean of a lognormal random variable with the log mean as $\ln S_0 + \mu T - \frac{1}{2}\sigma^2T$ and the log standard deviation as $\sigma \sqrt{T}$, and so the ...


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

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(...


4

The drift of $\mathrm{d}\ln(S_t)$ is indeed $r-\frac{1}{2}\sigma^2$ which is always negative if $r=0$. The extra $-\frac{1}{2}\sigma^2$ has many explanations. You could see it as a convexity correction (see Jensen's inequality) or martingale correction. Without it, $(S_t)$ wouldn't be a martingale. For the second part, note that $\ln(S_T)\sim N\left(\ln(S_0)...


4

Here are the things you need to correct in your code: Although you are setting a seed, you are generating the random numbers twice, and therefore they are not identical. Try this: rand = np.random.randn(N) y = np.exp(-volvol**2.0/2 * time_step + volvol * s * rand) y1 = rand You also need to multiply the sigma by the square root of the timestep in the SDE ...


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

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

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

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 $\Delta t$ each. 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

Let $Y = \log X$, then: $$\begin{align} Y &= Y_0 + (\mu-\frac{\sigma^2}{2})t + \sigma W_t \\ EY_t &=Y_0 + (\mu-\frac{\sigma^2}{2})t \\ EY_tEY_s &= Y_0^2 + Y_0 (\mu-\frac{\sigma^2}{2}) (t+s) + (\mu-\frac{\sigma^2}{2})^2 t s \\ E(Y_tY_s) &= E\left((Y_0 + (\mu-\frac{\sigma^2}{2})t + \sigma W_t) (Y_0 + (\mu-\frac{\sigma^2}{2})s + \sigma W_s)\...


3

Because $\mathbb{E}\left(e^{\sigma W_t}\right) = e^{\frac{1}{2}\sigma^2T} > 1$, you need that correction to ensure that your asset grows on average at rate $\mu$ (or $r$ in the risk-neutral measure). This is pretty well explained in the chapter on BS model from Hull’s book Options, futures and other derivatives!


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

Assuming that your GBM is given by $$S_{T}=S_{0}e^{(r -{\frac {\sigma ^{2}}{2}})T+\sigma W_{T}}$$ then its mean and variance are: $${Mean=S_{0}e^{r T},}$$ $$ {Variance=S_{0}^{2}e^{2r T}\left(e^{\sigma ^{2}T}-1\right)}{\displaystyle}$$ You cannot paste these values directly into np.random.lognormal because in this case the parameters $\mu$ and $\sigma^2$ ...


3

If $S_T<K_1$, the payoff is zero, and we have $\frac{(K_2-K_1)S(T)}{K_2} \geq0$ If $K_1 \leq S_T<K_2$, the payoff is $(S_T -K_1)$. We have $$K_1K_2 \geq S_TK_1$$ and $$S_TK_2+K_1K_2 \geq S_TK_2+S_TK_1$$ Thus, $$S_TK_2-S_TK_1 \geq S_TK_2-K_1K_2$$ Finally, $$S_T(K_2-K_1) \geq (S_T-K_1)K_2$$ $$\frac{S_T(K_2-K_1)}{K_2} \geq (S_T-K_1)$$ If $S_T \geq ...


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

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

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

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

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

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 ...


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