I'm currently trying to price some different kinds of contracts. I'm stuck on this following exercise, which I can't seems to find a good solution for. The following is assumed:

  • We are in a standard BS environment with $dS(t) = \mu S(t) dt + \sigma S(t) dW(t)$ with $\mu, \sigma > 0$.
  • Interest rate is $0$.
  • The $Q$ dynamics are: $dS(t) = \sigma S(t) dW(t)^{Q}$.
  • The payoff function is given as: $\left(\int_{0}^{T} \mathrm{e}^{a \cdot v}\ln(S(v))dv\right)^3$.

The task is to find the risk neutral valuation.

My approach was to start write the RNV function in our case, which is $$F(t,S(t))=\mathrm{e}^{-r(T-t)}E^{Q}[payoff] = E^{Q}\left[\left(\int_{0}^{T} \mathrm{e}^{a \cdot v}\ln(S(v))dv\right)^3\right]$$

Define $X(t) = ln(S(t))$, then applying Ito, we get $$\begin{align} dX(t) = \frac{1}{S(t)}dS(t) - \frac{1}{2S(t)^{2}}(dS(t))^2 &= \frac{1}{S(t)}(\sigma S(t) dW(t)^Q) - \frac{1}{2S(t)^{2}}(\sigma S(t) dW(t)^Q)^2\\ &= -\frac{1}{2}\sigma^{2}dt+\sigma dW(t)^{Q}\end{align}$$ Define now $Y(t) = \mathrm{e}^{a \cdot t}X(t)$. Applying Ito, we get $$\begin{align} dY(t) &= a\mathrm{e}^{a \cdot t}X(t)dt+\mathrm{e}^{a \cdot t}d(X(t)) =aY(t)dt+\mathrm{e}^{a \cdot t}(-\frac{1}{2}\sigma^{2}dt+\sigma dW(t)^{Q})\\ &=aY(t)dt-\mathrm{e}^{a \cdot t}\frac{1}{2}\sigma^{2}dt+\mathrm{e}^{a \cdot t}\sigma dW(t)^{Q} \end{align}$$ Integrating both sides we get $$ \begin{align} Y(T) &= Y(t) + (Z(T) - Z(t))-\mathrm{e}^{a \cdot t}\frac{1}{2a}\sigma^{2} + \sigma^{2}\int_{t}^{T} \mathrm{e}^{a \cdot t} dv \end{align} $$ where $Z(t) = \int_{0}^{t} Y(v) dv$.

This is where I'm not sure where to proceed or if my calculations up until now are correct.

  • 1
    $\begingroup$ I edited the reply, I now understand your initial point. $\endgroup$ Dec 16, 2021 at 14:35
  • $\begingroup$ Thank you for your detailed answer. This looks more correct to me. $\endgroup$
    – Marc Allan
    Dec 16, 2021 at 17:49

1 Answer 1


The proof strategy consists on showing the quantity of interest is normally-distributed, then using the moment-generating function of a normal variable to obtain its third moment.

Under measure $\mathcal{Q}$, we define \begin{align} \xi:&=\int_0^Te^{av}\ln S_v \text{d}v \\ &=\int_0^Te^{av}\left(\ln S_0-\frac{1}{2}\sigma^2v+\sigma W_v^\mathcal{Q}\right)\text{d}v. \end{align}

The mean $\mu$ of $\xi$ is equal to $$\mu:=\int_0^Te^{av}\left(\ln S_0-\frac{1}{2}\sigma^2v\right)\text{d}v.$$

Then $$\xi=\mu+\sigma\int_0^Te^{av}W_v^\mathcal{Q}\text{d}v,$$

Now per the stochastic Fubini theorem: \begin{align} \int_0^Te^{av}W_v^\mathcal{Q}\text{d}v &=\int_0^Te^{av}\left(\int_0^T1_{\{u\leq v\}}\text{d}W_u^\mathcal{Q}\right)\text{d}v \\ &=\int_0^T\left(\int_0^Te^{av}1_{\{u\leq v\}}\text{d}v\right)\text{d}W_u^\mathcal{Q} \\ &=\int_0^T\left(\int_u^Te^{av}\text{d}v\right)\text{d}W_u^\mathcal{Q} \\ &=\int_0^T\theta(u,T)\text{d}W_u^\mathcal{Q}, \end{align} where $$\theta(u,T):=\frac{e^{aT}-e^{au}}{a}$$

Yet we know that the stochastic integral above follows a Gaussian distribution, so $$\xi\overset{\mathcal{L}}{=}X,$$

where $$X\sim\mathcal{N}\left(\mu,\nu\right)$$ and $$\nu:=\sigma^2\int_0^T\theta(u,T)^2\text{d}u.$$ The moment-generating function $M(t)$ of a Gaussian random variable with mean $\mu$ and variance $\nu$ is $$M(t):=e^{\mu t+\frac{1}{2}\nu^2t^2}.$$ Differentiating 3 times: $$M^{\prime\prime\prime}(t):=\left(3\nu^2(\mu+\nu^2t)+(\mu+\nu^2t)^3\right)M(t).$$ Setting $t=0$ gives us the desired result: \begin{align} E(\xi^3) &= M^{\prime\prime\prime}(0) \\ &=3\nu^2\mu+\mu^3. \end{align}

  • $\begingroup$ Thank you a lot things definitely makes more sense now. I was wondering if your definition of $\xi$ is wrong. I'm not sure if we are allowed to take $\log(S_{0})$ out just like that. I can you elaborate for the justification of taking it out as a scalar. $\endgroup$
    – Marc Allan
    Dec 15, 2021 at 17:53
  • $\begingroup$ @MarcAllan as in any integral of the form $\int_D\alpha f(x)\text{d}x$, you can always take out elements of the integrand that do not depend on the variable of integration, i.e. $\alpha\int_Df(x)\text{d}x$. Here $\ln S_0$ is independent of time $v$. $\endgroup$ Dec 16, 2021 at 8:41
  • 2
    $\begingroup$ @DaneelOlivaw: if $\ln S_t = \ln S_0 - \frac{\sigma^2}{2}t + \sigma W_t$, shouldn't it be $\xi = \ln S_0\int_0^T e^{av}dv - \int_0^Te^{av}\left(-\frac{1}{2}\sigma^2v+\sigma W_v^\mathcal{Q}\right)\text{d}v$ instead of $\xi = \ln S_0\int_0^Te^{av}\left(-\frac{1}{2}\sigma^2v+\sigma W_v^\mathcal{Q}\right)\text{d}v$? $\endgroup$
    – Sebastian
    Dec 16, 2021 at 12:21
  • 1
    $\begingroup$ I agree with Sebastian, but one of the minus signs shouldn't be in Sebastians correction. $\endgroup$
    – Marc Allan
    Dec 16, 2021 at 12:38
  • 2
    $\begingroup$ @Sebastian you're right, I guess that's what OP also meant, apologies I didn't notice. I've edited accordingly. $\endgroup$ Dec 16, 2021 at 14:34

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