# 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:

1. Simulated 1000 Wiener processes (each with 1001 time steps between $t=0$ and $t=1$)
2. Based on these processes, I have created 1000 models for the evolution of the stock price $S_t$, based on the formula $$S_t = \exp(-W_t + t)$$
3. Based on the 1000 obtained values for $S_1$ I have calculated the payoff of the call option $C$ in each case
4. 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.

• what is your volatility? – Gordon Mar 22 '18 at 14:36
• I'm basing it off of a question which doesn't specify the volatility. I have therefore (somewhat arbitrarily) set it to be 1. – M Smith Mar 22 '18 at 14:39
• It appears that you switched $r$ and $T$. – Gordon Mar 22 '18 at 15:07
• Tag "homework" missing? – LocalVolatility Mar 22 '18 at 15:55
• I have added the tag – M Smith Mar 22 '18 at 16:14

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.

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.

• Thanks for your response. I have amended my post to show the exact question that I am attempting to answer. I have set volatility to $100\%$. – M Smith Mar 22 '18 at 15:24
• Maybe I am missing something but with $r=\sigma=1$ I don't see how your question reaches $S_t=S_0\exp(t-W_t)$. We have, under the implied probability measure: $S_t=S_0\exp((r-\sigma^2/2)t+\sigma W_t)=\exp((1-1/2)t+W_t)=\exp(t/2+W_t)$ $=\exp(t/2-W_t)$. – Daneel Olivaw Mar 22 '18 at 15:36
• $T$ should equal $1$ not $1\%$ in your question. – Daneel Olivaw Mar 22 '18 at 15:40
• Yes, that should have been the value of $r$ instead - a typo. – M Smith Mar 22 '18 at 15:42
• And r = 100%, not 1% (according to the pic saying use exp(-1)). I agree with Daneel that seems wrong (should have t/2). This is either a trick question (r=100%?), or a wrong question. – Yian Pap Mar 22 '18 at 16:13