# Exercise on arbitrage-free process

Consider the following problem, from Bjork's Arbitrage Theory in Continuous Time:

Consider the standard Black-Scholes model. Derive the arbitrage free price process for the $$T$$-claim $$\mathcal{X}$$ where $$\mathcal{X}$$ is given by $$\mathcal{X}=\{S(T)\}^\beta$$. Here $$\beta$$ is a known constant.

My approach.

Let $$F(t,s)$$ be the price of the claim $$\mathcal{X}$$ at time $$t$$, when the underlying spot price is $$s$$.

The Black-Scholes equation for $$F$$ is: \begin{align} F_t + rsF_s + \frac12 \sigma^2s^2 F_{ss} - rF &= 0 \\ F(T, S(T)) &= S(T)^\beta. \end{align}

It is convenient to make a change of variables of the form $$\tilde{F}(t,s) = e^{-rt}F(t,s)$$, so that the associated stochastic process is: \begin{align} dX &= rX dt + \sigma X dW \\ X(t) &= s. \end{align}

After changing variable to $$Y = \log X$$ and integrating, I find $$X(T) = s \exp\left((r-\frac12 \sigma^2)(T-t) + \sigma(W(T) - W(t))\right).$$ So by the Feynman-Kac formula I have: \begin{align} F(t,s) &= e^{-r(T-t)}\mathbb{E}\left[X(T)^\beta\right] \\ &= e^{-r(T-t)}\int_{-\infty}^{\infty}s^\beta e^{\beta z} \exp\left(-\frac12 \frac{(z - (r-\frac12\sigma^2)(T-t))^2}{\sigma^2(T-t)}\right) dz, \end{align} which after some computation gives, if I did not make any mistake: $$F(t,s)=e^{-r(T-t)}s^\beta \exp\left(\frac12\sigma^2\beta^2(T-t) + (r-\frac12\sigma^2)\beta(T-t)\right).$$

Does it sound right?

Also, regardless of whether the pricing formula is correct, I am not sure if what I found is really the arbitrage free stochastic process for $$\mathcal{X}$$.

If $$X$$ is a lognormally distributed variable, $$X = e^{\mu + \nu Z}$$, so $$\ln X$$ has mean $$\mu$$ and variance $$\nu^2$$, and $$Z$$ is normally distributed, then $$E\left[ X^n \right] = e^{n\mu + \frac{1}{2} n^2\nu^2}$$ This solves your question with $$X = S_T/S_t$$, $$n = \beta$$, $$\mu = (r -\frac{1}{2} \sigma^2) (T-t)$$ and $$\nu = \sigma \sqrt{T-t}$$ in the Black Scholes world.
• Ok, so using the expression for $E[X^n]$ as you suggest, and plugging in $X=S_T/s$, given that $s$ is deterministic I would have: $$F(t,s) = e^{-r(T-t)} s^\beta E[X^\beta],$$ which gives precisely the formula I found (I consider the discounted value at time $t$ of the risk-neutral expectation). Sep 11 '19 at 10:09