Simulating Black-Scholes Call Values Under the Risk-Neutral Measure

# Cause of difference in theoretical vs observed value of a (call) option under the Black-Scholes model?


# Cause of difference in theoretical vs observed value of a (call) option under the Black-Scholes model?












I am currently considering the price $C_0$ of a call option on a stock $S$ with $$ S_0 = 1 \\ K = 1.1 \\ r = 1\% \\ T = 1 $$

Based on the Black-Scholes formula, I have deduced that $C_0 = 0.356$.

However, I am currently trying to replicate this result in R. To do this, I have:

- Simulated 1000 Wiener processes (each with 1001 time steps between $t=0$ and $t=1$)

- Based on these processes, I have created 1000 models for the evolution of the stock price $S_t$, based on the formula $$ S_t = \exp(-W_t + t) $$

- Based on the 1000 obtained values for $S_1$ I have calculated the payoff of the call option $C$ in each case

- Discounting each of these values, by multiplying each by $e$, I have found 1000 possible values for $C_0$, which I have then taken the mean of

However, this method gives a result (of approx. 4) which varies significantly from the theoretical result obtained using the Black-Scholes formula. I am presuming that this is due to an error in my method used in R.

Can anyone help me to understand where I might be going wrong?

EDIT: The following image shows the exact question that I am attempting to answer.

## Answer by Ivan (score 2, accepted)

https://quant.stackexchange.com/a/39009

A curious piece of homework, but let’s just consider the information at hand.

You are given a somewhat odd process $S_t = S_0e^{-W_t+t}$ and the rest of the question pointing you to the “implied probability measure” looks like an indication that the stated process is under the physical measure $P$.

The SDE followed by $S$ under $P$ must be:

$\frac{dS_t}{S_t} = \frac{3}{2}dt - dW_t$ for any of this to make sense.

Under the risk-neutral measure $Q$ (and I’ll make the assumption this is what is meant by implied measure), we’ll have (reversing the sign of $dW_t$ for sanity)

$\frac{dS_t}{S_t} = dt + dW_t$

Since $r\equiv{1}$

And the solution to this is $S_T = S_0e^{\frac{1}{2}T+W_T}$

This is what you need to simulate $N$ times (you don’t need the full path) to compute the call payoff in one year, and from that its discounted expected value.

Note: the expected value (under $Q$) of $S_T$ here is $e^{1} = 2.71...$ and so intuitively your call of strike 1.1 is going to be very much in the money, with an undiscounted expected value of the order of 1.6 + some time value since the volatility is still pretty high ($\sigma=1$). Your discount factor is $e^{-1}=0.36...$ so your call should be worth something around 0.60. If you plug $S_0, T, r,\sigma, K$ into a BS pricer, that’s what you should find.

## Answer by Daneel Olivaw (score 0)

https://quant.stackexchange.com/a/38936

If $r=1$ and $\sigma=\sigma^2=1$, then: $$ S_t=\exp\left(\color{red}{\frac{t}{2}}-W_t\right)$$

I am a bit puzzled because you should be obtaining lower values than $C_0=0.356$ instead of higher. Also, do you mean $\sigma=1=100\%$ or $\sigma=1\%=0.01?$ $r=1=100\%$ or $r=1\%=0.01?$

Also, to avoid confusion, although $W_t$ and $-W_t$ have same distribution I would write: $$ S_t=\exp\left(\color{red}{\frac{t}{2}} \color{red}{+} W_t\right)$$

Finally, given this a European option, you can directly simulate the terminal state $S_1$ instead of the whole path between $S_0$ and $S_1$.

Let me know if this helps.