martingale pricing for european option question

I am taking a quant finance course, and am confused about my professor's methodology about the martingale method for pricing a standard European call, with payoff:

$c(T) = max[S(T) - K, 0]$,

which we write in terms of the martingale, $\hat{S}$ as:

$c(T) = exp(rT) max[\hat{S}(T) - exp(-rT)K,0]$

Then, he writes that:

$c(0) = \mathbb{E}_{\hat{\mathbb{P}}} \left[max [\hat{S}(T) - exp(-rT)K,0]\right]$

We can then write:

$c(0) = \int \left\{S(0) \exp[-1/2\sigma^2 T + \sigma x] - exp(-rT)K\right\} \times \frac{1}{\sqrt{T}\sqrt{2\pi}} \exp \left[-1/2 (x/\sqrt{T})^2\right]dx$.

I understand everything except the multiplication by the Brownian density: $\exp \left[-1/2 (x/\sqrt{T})^2\right]$. The reason is I thought this expectation is taken with respect to $\hat{\mathbb{P}}$ and not $\mathbb{P}$, where it is my understanding/confusion that this new probability density is obtained via the Radon-Nikodym derivative:

$\frac{d\hat{P}}{dP} = \exp\left[-\left(\frac{\mu -r}{\sigma}\right)W(t) - \frac{1}{2}\left(\frac{\mu -r}{\sigma}\right)^2 t\right]$ , so shouldn't the expectation have this extra term in it? I.e.,

$\hat{\mathbb{P}} = \mathbb{P} \times \frac{d\hat{P}}{dP}$?

Thanks!

The change of measure was done beforehand, $\hat{S}$ is a martingale under the $\hat{\mathbb{P}}$-measure, hence driftless. In the standard context that your professor adopted, $S$ is a log-normal process.

Therefore, one can write $$\hat{S(T)}=\hat{S(0)}e^{-1/2\sigma^2 T + \sigma W(T)}$$ where $W$ is a $\hat{\mathbb{P}}$-standard Brownian motion.

By replacing this formula in your professor's, we have

$$c(0) = \mathbb{E}_{\hat{\mathbb{P}}} \left[max [\hat{S(0)}e^{-1/2\sigma^2 T + \sigma W(T)}- exp(-rT)K,0]\right]$$

The only source of randomness in your formula is $W(T)$( centered,normally distributed with variance $T$) and it happens that we know its density function which is $$f_{W(T)}(x)=\frac{1}{\sqrt{T}\sqrt{2\pi}} \exp \left[-1/2 (x/\sqrt{T})^2\right]$$ Finally,

$$c(0) = \mathbb{E}_{\hat{\mathbb{P}}} \left[max [\hat{S(0)}e^{-1/2\sigma^2 T + \sigma W(T)}- exp(-rT)K,0]\right]=\int_{\mathbb{R}}{max [\hat{S(0)}e^{-1/2\sigma^2 T + \sigma x}- exp(-rT)K,0]f_{W(T)}(x)dx}$$

which is the result you are looking for ( be careful, you forgot the max function in your integral)

• Hi. That makes sense, thanks. So, in general, if your only source of randomness is W(T), no matter what the Radon-Nikodym derivative is, you can always just integrate with $f_{W(T)}$? – Thomas Moore Sep 6 '17 at 22:22
• In the answer, we are only working under one measure, so we could do that. However, if $W$ was a $\hat{\mathbb{Q}}$ Brownian motion, either you use your Radon-Nikodym derivative to change the expectation measure to $\hat{\mathbb{P}}$, or you convert $W$ into a $\hat{\mathbb{P}}$-measurable process. For example, we could have defined $S$ as $$dS_t=S_t(\mu dt+\sigma dW_t^{h})$$ where $W^{h}$ is a Brownian process under a measure $\mathbb{H}$, in that case we have to change the measure so it is compatible with the measure of the expectation, which is $\hat{\mathbb{P}}$ here. – Canardini Sep 7 '17 at 12:58