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 $\mathcal{X}$.


1 Answer 1


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.

See also the following wiki page for other properties of lognormal distribution:


  • $\begingroup$ 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). $\endgroup$
    – J. D.
    Sep 11, 2019 at 10:09
  • $\begingroup$ @J.D. yes, that's correct. It's good/useful that you derived it yourself. $\endgroup$
    – user34971
    Sep 11, 2019 at 10:50

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