Arbirtage free price process question in Bjork's Arbitrage Theory in Continuous Time

Question

I am currently working through questions in Bjork's Arbitrage Theory in Continuous Time. However, I am unable to solve the following question, 7.2 in the book. A solution would be greatly appreciated.

Consider the Black Scholes model. A company has produced the derivative the Golden Logarithm, henceforth abbreviated as the GL. The holder of the GL with maturity time T, denoted as GL(T), will, at time T, obtain the sum lnS(T). Note that if S(T)<1 this means that the holder has to pay a positive amount to the company. Determine the arbitrage free price process for the GL(T)

Answer 1

In general, the arbitrage-free price process $V_t$ at time $0 \le t \le T$ for a European claim $X =f(S_T)$ under the B-S model (which it looks like you have) is given by

$$V_t(X) = B_t\mathbb{E}_\mathbb{Q}[B_T^{-1}X | \mathcal{F}_t],$$

where $B_t$ is the bond price process, $\mathbb{Q}$ is the measure making the discounted B-S stock price process a $\mathbb{Q}$-martingale, and $\mathcal{F}_t$ is the sigma field for the Brownian Motion. This might be the "process" the question asks for, but you would need to evaluate this expectation at each $t$ to actually use it.

Here's an example of computing the initial value using an arbitrary payoff at maturity, $X = f(S_T)$, so all you'll have to do is replace the $f$ with your logarithm payoff.

I assume you're familiar with the B-S stock price process under the risk-neutral Brownian motion, $S_t = S_0\exp(\sigma \widetilde{W}_t + (r - \frac{1}{2}\sigma^2)t)$, where $\widetilde{W}_t$ is a $\mathcal{N}(0, t)$-distributed random variable for fixed $t$ under the $\mathbb{Q}$ measure, $r$ is the risk-free rate and $\sigma$, the volatility. Note the argument of the exponential function is a $\mathcal{N}((r - \frac{1}{2}\sigma^2)t, \sigma^2 t)$-distributed random variable for fixed $t$.

Define the random variable $Y$ ~ $\mathcal{N}(-\frac{1}{2}\sigma^2 T, \sigma^2 T)$. Then $S_T = S_0\exp(Y + rT)$, and letting $p(Y)$ denote the pdf of $Y$, from the formula for $V_t$ we have

$$V_0 = \mathbb{E}_\mathbb{Q}[\mathrm{e}^{-rT}f(S_0\exp(Y + rT))] \\ = \mathrm{e}^{-rT} \int_{-\infty}^\infty f(S_0\exp(y + rT))p(y)\mathrm{d}y \\ = \mathrm{e}^{-rT} \int_{-\infty}^\infty f(S_0\exp(y + rT))\frac{1}{\sqrt{2 \pi \sigma^2 T}}\exp\left(\frac{-(y + \frac{1}{2}\sigma^2T)^2}{2 \sigma^2 T}\right)\mathrm{d}y.$$

Answer 2

I am also doing the same exercise. Let $GL(t,s)$ denote the value of the Golden Logarithm at time $t$ when the underlying stock has price $s$.

As in the answer by bcf, the Black-Scholes model gives: $$ GL(t,s) = e^{-r(T-t)}\mathbb{E}^Q[GL(T)]=e^{-r(T-t)}\mathbb{E}^Q[\log S(T)], $$ with expectation taken over the risk-free measure.

The stochastic process underlying the Black-Scholes model is: $$ \begin{align} dS &= rSdt + \sigma SdW \\ S(t) &= s \end{align}, $$ so that the stochastic process for the golden logarithm is: $$ \begin{align} d(GL)&=\left(r-\frac12 \sigma^2\right)dt + \sigma dW \\ GL(t)&=\log s \end{align}. $$ This process can be immediately integrated to: $$ GL(T) - \log s = \left(r-\frac12\sigma^2\right)(T-t) + \sigma\left(W(T)-W(t)\right). $$ Replacing this in the expectation above, we have $$ GL(t,s) = e^{-r(T-t)}\int_{-\infty}^\infty \left[\log s + \left(r-\frac12 \sigma^2\right)(T-t) + z\right] f(z) dz, $$ where $f(z)$ is the normal density with mean 0 and variance $\sigma^2(T-t)$. From this expression we can immediately conclude: $$ GL(t,s) = e^{-r(T-t)}\left[\log s + \left(r-\frac12\sigma^2\right)(T-t)\right]. $$ I hope this is right, but I am not sure as I am also learning this stuff.