I am facing a question that I thought I was doing right, but I've become skeptical since the European Put price is crazy. I know this is a finance question, but I feel like I'm probably screwing up the math involved. Anyways, I'm asked

Assume that the risk free rate r is 0 and that the stock price is given by the equation $$ S(t)=8e^{2t+2B(t)} $$ where $B(t)$ is the standard Brownian motion. Determine the price at time 0 of the European call option and European put option with strike $K=8e^{18}$ and expiration t = 9.

What I know... I think

I know that the format of the formula for $S(t)$ is $$ S(t)=S_0e^{\mu t + \sigma B(t)} $$ so $\mu = 2$, $\sigma = 2$, and so $\sigma^2 = 4$.

I also have the formula for the price of a European Call Option, $C_0$ $$ C_0=S_0 \phi(d_+) -Ke^{-rt}\phi(d_-) $$ where $$ d_\pm = \frac{ln(\frac{S_0}{K})+(r \pm \frac{\sigma^2}{2})t}{\sigma \sqrt{t}} $$ and $$ \phi(x)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{x}e^{-\frac{z^2}{2}}dz $$

The Work I Have Done

From the information given, I got that $d_+ = 0$ and $d_- = -6$. Performing $\phi()$ on both ds gave me $$ \phi(d_+) = \phi(0) = \sqrt{\frac{\pi}{2}}*\frac{1}{\sqrt{2\pi}} $$ and $$ \phi(d_-) = \phi(-6) = 2.47301*10^{-9}*\frac{1}{\sqrt{2\pi}} $$

Substituting these into the $C_0$ equation, along with the constants given in the question, yielded $$ C_0=S_0 \phi(d_+) -Ke^{-rt}\phi(d_-)=8\phi(0)-8e^{18}e^{-0*9}\phi(-6)\\ =8\sqrt{\frac{\pi}{2}}*\frac{1}{\sqrt{2\pi}}-8e^{18}*2.47301*10^{-9} *\frac{1}{\sqrt{2\pi}}\\ = 4-8e^{18}*2.47301*10^{-9}*\frac{1}{\sqrt{2\pi}}\\ = $3.48 $$

Where the Problem Arises I Think

This price for the call is a bit higher than the typical examples I do, but it seems reasonable enough. The big problem I am having is that when I try to do the European Put Option, using the Call-Put Parity $$ C_0-P_0=S_0-Ke^{-rt} \rightarrow P_0=C_0-S_0+Ke^{-rt} $$ where $P_0$ is the price of the put option, the price I find is incredibly high due to that $K=8e^{18}$. Furthermore, using the actual formula for the European Put Price $$ P_0=Ke^{-rt}\phi(-d_-)-S_0 \phi(-d_+) $$ Leaves an incredibly high European Put Price too.

Any help you can provide on this question would be greatly appreciated. Sorry for how long this post is and thanks for the assistance!

  • 1
    $\begingroup$ Why do you think you're wrong? $\endgroup$
    – Mark Viola
    Commented Dec 13, 2017 at 20:34
  • $\begingroup$ Just how large a price I'm calculating for the put option makes me very skeptical @MarkViola $\endgroup$
    – strwars
    Commented Dec 13, 2017 at 22:30

1 Answer 1


The correct call price is $C_0 = 3.48$ and the correct put price is $P_0 = 525,279,748.58$ to two decimal places.

The strike price $K = 8e^{18} \approx 525,279,753$ is incredibly high, given that the initial stock price is $S_0 = 8$. Granted the volatility of $200 \,\%$ and the time-to-expiration of $9$ years are relatively large, but the expected future stock price grows with a drift equal to the risk-free rate $r = 0$, not $\mu = 200 \, \%$, under the risk-neutral measure -- and this is the only measure that matters to the option price.

Consequently, the probability that the stock price exceeds the strike at expiration (in the risk-neutral world) is very small. The out-of-the-money call price tends to $0$ as $K \to \infty$, but the in-the-money put price tends asymptotically to the intrinsic value $K- S_0$.

The call and put prices you find are not contradictory.

  • $\begingroup$ @Mark Viola: Thanks very much. Not much of a market for $10^{16}$ % in the money puts. $\endgroup$
    – RRL
    Commented Dec 13, 2017 at 23:34
  • $\begingroup$ That's a pretty deep ITM put. $\endgroup$
    – Mark Viola
    Commented Dec 14, 2017 at 0:14

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