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I have a question considering Financial markets in discrete Time:

One of the main theorems in discrete time is:

In finite discrete Time with trading times t={1,...,T} the following are equivallent:

1.)The market $(S,\mathbb{F})$ is complete, i.e every $F_T$-measurable random variable $U$ is replicable
and

2.)there is exactly one equivalent martingale measure.

Now I seem to come to a contradiction if I define the following frame work (shortly: Black scholes in discrete time):

Lets assume we are in the Black-Scholes framework but we consider the model as discrete time model with (simplifying heavily) two trading dates $t=0$ and $t=1$. The (discounted) Stock price ist denoted by $S_t$. $S_0$ is constant and for T=1 $$S_T=exp((\mu-r-1/2\sigma^2)T-\sigma W_t).$$ The Filtration $\mathbb{F}=(F_t)_{t=0,1}$ is the natrual one.

It is clear that in this frame work there is exactly one equivalent martingale measure, namely the one in whick $\mu=r$. Applying the theorem from above a call-option on that stock S should be replicable (in only one! trading day, namely from t=0 to T=1)

Now this seems to me very doubtfull and I dont really know where the Problem is.... ANY HELP IS WELCOME!!!!

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Hi Bob, welcome to quant.SE. To get a good answer to your question, it would help a lot if you worked on improving the wording and formatting of your question. I've gotten you started, here, but take a look around the site at popular questions to get a sense for the level we expect here. –  Tal Fishman Oct 7 '11 at 11:39
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2 Answers

A discrete-time model only works in no-arbitrage land with discrete asset values. Furthermore, the number of allowable asset values per timestep is limited by the number of available securities.

The tree is the classic example of this. Binomial trees "work", but if you make a one-step trinomial tree, you will find that you can no longer form a risk-free portfolio from an option and its underlying.

(Of course, as a way of numerically solving the PDE, trinomial trees are still fine.)

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Exact Discretization of the Solution to the Geometric Brownian Motion Stochastic Differential Equation

Let $P_{t}$ represent the time series of market prices of the underlying, $\mu$ be its mean continuous log-return, $\sigma$ be its instantaneous volatility and $W_{t}$ be a Wiener process.

Here is the stochastic differential equation for the geometric Brownian motion:

$$ \frac{dP_{t}}{P_{t}}=\mu dt+\sigma dW_{t} $$

Here is the exact solution to the equation:

$$ \ln\left(\frac{P_{t}}{P_{0}}\right)=\left(\mu-\frac{\sigma^{2}}{2}\right)t+\sigma W_{t} $$

The discretization of this solution over a small but finite interval $\delta$ is given by the following:

$$ \ln\left(\frac{P_{t+\delta}}{P_{t}}\right)=\left(\mu-\frac{\sigma^{2}}{2}\right)\delta+\sigma W_{t} $$

where $W_{t}$ amounts to a standard normal variate $Z_{t}$ times the square root of the time interval $\delta$, so that $W_{t}=Z_{t}\sqrt{\delta}$.

If the time to maturity is $T$, the number of time steps corresponding to the time interval $\delta$ is given by $n=\frac{T}{\delta}$. Thus,

$$ \ln\left(\frac{P_{T}}{P_{0}}\right)=\left(\mu-\frac{\sigma^{2}}{2}\right)T+\sigma\sqrt{\delta}\sum_{k=1}^{n}Z_{k} $$

When one perform simulations, a path is represented by the foregoing formula, but there are as many instances of this formula as there are paths to simulate, so that even if the deterministic part of the formula is the same from path to path, the stochastic part of the formula, the $Z_{k}$, have to be generated anew for each instance of the formula, so that there are n times the number of simulations standard normal random numbers to generate in order to generate one path per simulation. Of course, the bigger the number of simulated paths and the smaller the $\delta$, the more realistic are the results of the simulation.

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Ah, excellent. I've deleted your original posted link so just this would stand out. And yes, the $\LaTeX$ shows-up just fine in my browser. –  chrisaycock Jan 16 '12 at 19:50
    
You have discretized the path-generating part, but the existence of the equivalent martingale measure that bob asks about still requires more structure. –  Brian B Jan 17 '12 at 13:57
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