Earlier I had the question (5.11 Tomas Bjork):
$$ \frac{\partial F}{\partial t}+\frac{1}{2}x^2\frac{\partial^2 F}{\partial t^2}+x = 0 $$ $$ F(T,x) = ln(x^2) $$ And solve it using Feynman-Kac. The PDE gives the stochastic differential $dX=XdW$ so to solve the PDE I need what $X$ that solves the stochastic differential. An example in the book showed how to solve a differential of the type above and did it by letting $Z=lnX$, compute its differential $dZ$ (which then become independent of Z, which I guess is the point), solve for Z, and then for X just raise the exponential function with the solution for Z (since $Z=lnX$). This all went fine.

Now I have a question considering the standard BS-model where I´m supposed to derive the arbitrage free proce process for the contingent claim $X=(S(T)^{\beta})$, $\beta=const.$

Standard BS-model gives the differential for $S$ which is $dS=rSdt+\sigma S dW$. Thinking of the earlier question I gave it a try doing as before. Setting $Y=S^{\beta}$, and computing $dY$ gives:
$dY = Y(r\beta +\frac{\beta(\beta-1)}{2}\sigma^2)dt+\sigma \beta Y dW$
for which I then need the $Y$ that solves this $dY$. Doing as above however, setting $Z=ln Y$ (with $Z_0 = ln s_0$), computing $$dZ = \frac{1}{Y}dY-\frac{1}{2Y^2}(dY)^2 = dt(\beta r+\frac{\beta(\beta-1)}{2}\sigma^2-\frac{\beta^2\sigma^2}{2})+\beta \sigma dW$$ Integrating and raising the exponential function would give the solution $Y$ for $dY$: $$ Y = s_0 exp(\int^T_t(\beta r+\frac{\beta(\beta-1)}{2}\sigma^2-\frac{\beta^2\sigma^2}{2})dt + \int^{W_T}_{W_t}\beta \sigma dW)$$ which then would give the price process $$\pi = s_0^\beta exp[\beta(\beta r+\frac{\beta(\beta-1)}{2}\sigma^2-\frac{\beta^2\sigma^2}{2})(T-t)]$$ This result, according to the solutions manual is wrong however. It has this extra term $\frac{\beta^2\sigma^2}{2}$ which appears when doing the extra step of letting $Z=lnY$ and so on but I can´t understand what the difference is from the first case and why this doesn´t work. Please let me know if something is unclear and I´ll try to revise.


1 Answer 1


You're making it much more complicated than it has to be. Remember that the arbitrage free price is given by $e^{-r(T-t)}\mathbb{E}(\Phi(S_T))$ (to be correct, this should be conditioned on the appropriate sigma-algebra. I'll exclude that for ease of writing). Now use Itos on $\ln(S_t) = X_t$, this yields, after integration (Note that you integrate from $t$ to $T$ in all integrals. I see that you've written the integral from $W_t$ to $W_T$, this is incorrect):

$S_T = S_t \cdot \text{exp}((r-\sigma^2/2)(T-t) + \sigma Z$ where $Z \in N(0,\sqrt{T-t})$.

Now insert this into the expression for the price:

$e^{-r(T-t)}\mathbb{E}(S_T^{\beta})=S_t^{\beta}e^{-r(T-t)+\beta(r-\sigma^2/2)(T-t)}\mathbb{E}(e^{\beta Z})=S_t^{\beta}e^{-r(T-t)+\beta(r-\sigma^2/2)(T-t)+\beta^2\sigma^2(T-t)}$

This you can rewrite as you'd like.

As I mentioned previously. Note that you integrate from $t$ to $T$ and you'll get:

$X_T - X_t = \int_t^T (...)du + \int_t^T(...)dW_u$. Writing the last integral with limits $W_t$ to $W_T$ is incorrect!

Hope this makes it clear to you! This is the usual way to price contingent claims, using the PDE directly is only necessary when you can't compute the price analytically and must approximate it using numerical methods (you can use Monte-Carlo simultation also, for this the PDE is not required either).

  • $\begingroup$ Hehe, kind of what I guessed I did but wanted to try to do it from scratch to get a better notion of what I was doing. Not that it worked.. Anyhow, so when having a claim with some function based on a regular, say stock, there´s never need to compute any new differentials based on the given function? $\endgroup$ Apr 12, 2014 at 22:40
  • $\begingroup$ Hmm, it might be necessary in some cases. It worked nicely now since you took the power of an exponential, so you could just multiply the $\beta$ into the exponent and then the expectation became very easy. If this is not the case you should try to find the dynamics "of the whole claim" instead. $\endgroup$ Apr 13, 2014 at 9:33
  • $\begingroup$ Allright, but would the method for that be the way as I tried or would it contain something more in order to solve the new differential (i.e. the differential for the whole claim)? $\endgroup$ Apr 13, 2014 at 15:14
  • $\begingroup$ Just to be certain here, is $E[e^{\beta \sigma z}]=e^{\beta^2 \sigma^2 (T-t)}$? Because otherwise I dont get how the last part comes in. @GoodGuyMike $\endgroup$ Apr 22, 2014 at 15:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.