Why Black-Scholes Replication Fails with Continuous Outcomes in One Step

# Why doesn't Black-Scholes work in discrete time?


# Why doesn't Black-Scholes work in discrete time?












I have a question considering Financial markets in discrete Time.

One of the main theorems in discrete time is the following. In finite discrete Time with trading times t={1,...,T} the following are equivalent:

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

- There is exactly one equivalent martingale measure.

Now I seem to come to a contradiction if I define the following framework (shortly: Black Scholes in discrete time):

Let's 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 is 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 natural one.

It is clear that in this framework there is exactly one equivalent martingale measure, namely the one in which $\mu=r$. 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 doubtful and I don't really know where the Problem is.... ANY HELP IS WELCOME!!!!

## Answer by Brian B (score 6)

https://quant.stackexchange.com/a/2136

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.)

## Answer by Jean-Victor Côté (score 3)

https://quant.stackexchange.com/a/2782

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.