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Is there a closed form solution of the following price formula? Assuming $dS_t=rSdt+\sigma S_t dW_t$ under the Q dynamics

$e^{-r(T-t)}\mathbb{E}_t^\mathcal{Q}[(\frac{(\int_0^T S_u du)}{T})^2]$

I know that the integral of a geometric brownian motion has no nice distribution, but is it the same with the square of the integral?

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3 Answers 3

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Yes it is known in closed form. See https://www.rocq.inria.fr/mathfi/Premia/free-version/doc/premia-doc/pdf_html/asian_doc/asian_doc.html section 5.1 which references an older Geman-Yor paper.

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  • $\begingroup$ This gives a closed form for the Laplace transform of the Asian call option. Computing the second moment is a much simpler task. $\endgroup$
    – AFK
    Mar 17, 2015 at 18:59
  • $\begingroup$ Agree the moment calc is simpler than laplace transform. The reference I linked gives both; equation for moments is (11) in section 5.1. $\endgroup$
    – q.t.f.
    Mar 17, 2015 at 19:10
  • $\begingroup$ My bad. Good link. $\endgroup$
    – AFK
    Mar 17, 2015 at 19:40
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A few tips. First note that $e^{-rt}S_t$ is a martingale. So make it appear and then integrate by part to rewrite $\int S_u du$ as a stochastic integral. Finally use the Ito isometry property.

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The $P$ dynamics of the underlying asset are: \begin{align*} dS=S(\mu dt+\sigma dB_t) \end{align*} That has the following solution under the $\mathcal{Q}$ dynamics: \begin{align*} S_t=S_0 e^{(r-\frac{\sigma^2}{2})t+\sigma W_t} \end{align*} Where $W_t$ is the equivalent martingale with respect to the original geometric brownian motion. Define $Y_t=\int_0^t S_u du$, then according to Feynman-Kac the value of the replicating portfolio is given by \begin{align*} V_t&=e^{-r(T-t)}\frac{1}{T^2}\mathbb{E}_t^\mathcal{Q}[(Y_t+\int_t^T S_udu)^2)\\ &=\frac{e^{-r(T-t)}}{T^2} \mathbb{E}_t^\mathcal{Q}[Y_t^2 + 2Y_t \int_t^T S_u du + (\int_t^T S_u du)^2]\\ \end{align*} So: \begin{align*} \mathbb{E}_t^\mathcal{Q}[\int_t^T S_u du]&=S_t \mathbb{E}_t^\mathcal{Q}[\int_t^T \frac{S_u}{S_t} du]\\ &=S_t \int_t^T e^{(r-\frac{\sigma^2}{2})(u-t)}\mathbb{E}^\mathcal{Q}[e^{\sigma (W_u-W_t)}] du\\ \end{align*} Since $W_u-W_t$ follows $\mathcal{N}(0, u-t)$ under $Q$. According to the m.g.f. \begin{align*} &=S_t \int_t^T e^{r(u-t)}du\\ &=\frac{S_t}{r}(e^{r(T-t)}-1)\\ &=\frac{x}{r}(e^{r(T-t)}-1) \end{align*} Lets compute now the expectation of the square of the integral: \begin{align*} \mathbb{E}_t^\mathcal{Q}[(\int_t^T S_u du)^2]&=S_t^2 \mathbb{E}_t^\mathcal{Q}[\int_t^T \frac{S_u}{S_t}du \int_t^T \frac{S_v}{S_t}dv ]\\ &= S_t^2 \int_t^T \int_t^T \mathbb{E}^\mathcal{Q}[\frac{S_u}{S_t} \frac{S_v}{S_t}]dv du\\ \end{align*} Lets focus on: $\mathbb{E}^\mathcal{Q}[\frac{S_u}{S_t} \frac{S_v}{S_t}]$, lets assum $t \leq v \leq u$ \begin{align*} \mathbb{E}^\mathcal{Q}[\frac{S_u}{S_t} \frac{S_v}{S_t}&=\mathbb{E}^\mathcal{Q}[e^{(r-\frac{\sigma^2}{2})(u-t)+\sigma(W_u-W_t))}e^{(r-\frac{\sigma^2}{2})(v-t)+\sigma(W_v-W_t))}]\\ &=e^{(r-\frac{\sigma^2}{2})(u-t)+(r-\frac{\sigma^2}{2})(v-t)}\mathbb{E}^\mathcal{Q}[e^{\sigma(W_u-W_v)+2\sigma (W_v-W_t)}]\\ &=e^{(r-\frac{\sigma^2}{2})(u-t)+(r-\frac{\sigma^2}{2})(v-t)}\mathbb{E}^\mathcal{Q}[e^{\sigma(W_u-W_v)}]\mathbb{E}^\mathcal{Q}[e^{2\sigma(W_u-W_t)}]\\ &=e^{(r-\frac{\sigma^2}{2})(u-t)+(r-\frac{\sigma^2}{2})(v-t)}e^{\frac{\sigma^2}{2}(u-v)}e^{2\sigma^2(v-t)}\\ &= e^{ur}e^{u(r+\sigma^2)}e^{-t(2r+\sigma^2)} \end{align*} So: \begin{align*} \mathbb{E}_t^\mathcal{Q}[(\int_t^T S_u du)^2]&=2S_t^2 \int_t^T \int_t^u e^{ur}e^{u(r+\sigma^2)}e^{-t(2r+\sigma^2)}dv du\\ &= \frac{2x^2}{r+\sigma^2}(\frac{1}{2r+\sigma^2}e^{(2r+\sigma^2)(T-t)}-\frac{1}{r}e^{r(T-t)}+\frac{r+\sigma^2}{(2r+\sigma^2)r}) \end{align*} Mixing everything together we get: \begin{align*} V(t,x,y)&=\frac{y^2}{T^2}e^{rt-rT}+\frac{1}{T^2}\frac{2xy}{r}(1-e^{rt-rT})\\ &+ \frac{1}{T^2}\frac{2x^2}{r+\sigma^2}(\frac{1}{2r+\sigma^2}e^{(r+\sigma^2)(T-t)}-\frac{1}{r}+\frac{r+\sigma^2}{(2r+\sigma^2)r}e^{-r(T-t)}) \end{align*}

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  • $\begingroup$ Are you sure that your derivation is correct? It doesn't look so to me. $\endgroup$
    – SmallChess
    Mar 19, 2015 at 23:24
  • $\begingroup$ I can check it, but it seems ok to me. It satisfies the SDE $V_t + rxV_x+\frac{1}{2}\sigma^2 x^2 V_{xx} +x V_y=rV$ and boundary conditions. Why it does not look correct to you? $\endgroup$
    – Juan Imbet
    Mar 20, 2015 at 23:41

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