16 votes
Accepted

From VG and NIG processes to GBM

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 ...
Kevin's user avatar
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15 votes
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Find a formula for the price of a derivative paying $\max(S_T(S_T-K),0)$

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 ...
Kevin's user avatar
  • 15.9k
8 votes

What the expectation of S^2 is from GBM?

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{...
Kevin's user avatar
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8 votes
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Dynamics of FX rate

I am answering now instead of commenting. The rate of change in FX is naturally forward looking in this case. What you confuse is what happened to Spot due to changes in interest rate environments ...
AKdemy's user avatar
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6 votes
Accepted

Simulation of Geometric Brownian Motion in R

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. (...
Kevin's user avatar
  • 15.9k
6 votes
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Probability of an Option maturing In-the-money vs. Volatility

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}\...
Jan Stuller's user avatar
  • 6,098
6 votes
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Pricing an Option with payoff $\left(1-\frac{K}{S_t}\right)^{+}$

$\frac{1}{S_t}$ 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*} \...
Kevin's user avatar
  • 15.9k
5 votes

What is the stock price expectation?

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 ...
Slade's user avatar
  • 656
5 votes
Accepted

Normality or Log-Normality of Regular Returns

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 (...
Kevin's user avatar
  • 15.9k
5 votes
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Proving $\mathbb{P}(S_t<0|S_0=s_0)=0$ for Geometric BM

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 ...
Jan Stuller's user avatar
  • 6,098
5 votes
Accepted

Default intensity in Black-Cox model

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}\...
ir7's user avatar
  • 5,043
4 votes
Accepted

How to Understand Lognormal Distribution in the Following Case

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 ...
Kevin's user avatar
  • 15.9k
4 votes
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Why do I get this difference when simulating geometric Brownian motion?

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: ...
David Duarte's user avatar
  • 5,815
4 votes

Find a formula for the price of a derivative paying $\max(S_T(S_T-K),0)$

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 ...
Kupoc's user avatar
  • 98
4 votes

Normality or Log-Normality of Regular Returns

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 $\...
stackoverblown's user avatar
4 votes

Probability of an Option maturing In-the-money vs. Volatility

This is quite a brain teaser, at least it was for me. The way I thought about this initially was based on statistics. This lead me to believe that higher IVOL should always decrease the probability of ...
AKdemy's user avatar
  • 8,739
4 votes
Accepted

Geometric Brownian Motion simulation in Python: strange results

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 ...
Pontus Hultkrantz's user avatar
4 votes
Accepted

How to compute the Present Value of this path-dependent option?

We have $$ \begin{align} V(t) &= \mathbb{E}^{Q}[\mathbb{1}_{S(T_1)>B} (S(T_2)-K)^+)] \\ &= \mathbb{E}^{Q}[\mathbb{1}_{S(T_1)>B}\mathbb{1}_{S(T_2)>K} (S(T_2)-K))] \\ &= \mathbb{E}^{...
NN2's user avatar
  • 1,008
3 votes

What is the meaning that Geometric Brownian motion is leptokurtic?

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 ...
John's user avatar
  • 370
3 votes
Accepted

Compute the price of a derivative which pays $\log(S_T)S_T$ in the Black Scholes world

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. ...
Kevin's user avatar
  • 15.9k
3 votes

Compute the price of a derivative which pays $\log(S_T)S_T$ in the Black Scholes world

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=...
Jan Stuller's user avatar
  • 6,098
3 votes

Find a formula for the price of a derivative paying $\max(S_T(S_T-K),0)$

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 & ...
Jan Stuller's user avatar
  • 6,098
3 votes

Find a formula for the price of a derivative paying $\max(S_T(S_T-K),0)$

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 "...
alexprice's user avatar
  • 861
3 votes
Accepted

Covariance of logarithms of geometric Brownian motion

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+...
Jónás Balázs's user avatar
3 votes
Accepted

How To Understand the Drift of ln(S) if S Follows Geometric Brownian Motion

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)...
siou0107's user avatar
  • 2,680
3 votes
Accepted

Probability of a stock price using implied volatility

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 ...
Jan Stuller's user avatar
  • 6,098
3 votes
Accepted

Drawing values from a lognormal distribution of a GBM

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^{\...
emcor's user avatar
  • 5,795
3 votes
Accepted

Boundaries for Call Spread

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 ...
Canardini's user avatar
  • 553

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