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This is something that keeps bothering me about the Benchmark approach of Platen, which (very) shortly is as follows: Compare the development of an economic value with a growth optimal portfolio. Taking the expectation in the real world measure $\mathbb{P}$, conditioned on today's information, this should yield a fair and in some sense arbitrage free price, because you compare to the best possible (in expectation). Now, I want to know the price of a future on $X_T$ (hence $F(T,T)=X_T$). In discrete time, one just has to take the discounted future payments. For a time interval partition $\sigma$, this yields:

$$\Sigma_t^{(\sigma)} = \sum_{i=1}^N\frac{1}{S^{\pi^*}_{t_i}}( F(t_i,T) - F(t_{i-1},T) )\\ = {\sum_{i=1}^N\frac{1}{S^{\pi^*}_{t_{i-1}}}( F(t_i,T) - F(t_{i-1},T) )} +{\sum_{i=1}^N\left(\frac{1}{S^{\pi^*}_{t_i}}-\frac{1}{S^{\pi^*}_{t_{i-1}}}\right)( F(t_i,T) - F(t_{i-1},T) )}$$

Going to continuous time, this converges to $\int_t^{T} \left(\frac{1}{S^{\pi^*}_{s-}}\right) dF(s,T)+\int_t^{T}d\left[\left(\frac{1}{S^{\pi^*}}\right),F\right]_s$. Since the net position is zero and using the product rule for semimartingales, we get for the futures price at time t: $$ \mathbb{E}\left[ \frac{F(T,T)}{S^{\pi^*}_{T}}-\int_t^{T}F(s-,T)d\left( \frac{1}{S^{\pi^*}_s} \right) \big|\mathcal{F}_t\right] = \mathbb{E}\left[ \frac{F(t,T)}{S^{\pi^*}_{t}} \big|\mathcal{F}_t\right] = \frac{F(t,T)}{S^{\pi^*}_{t}} $$ To me it seems intractable. Is there any way to come the futures price process even close? I read some extensions to the usual equivalent martingale measure, cont., FV interest process one takes usually and which yields $F(t,T)=\mathbb{E}^{\mathbb{Q}}[F(T,T)|\mathcal{F}_t]$. But none of these can be applied here. Any ideas?

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@ user13655 : I think this question should be displaced to the quant.stackexchange.com forum. Best regards. –  TheBridge May 10 '12 at 15:38
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1 Answer

Well, at least there is a solution for deterministic interest rate. As an example take a BS model:

$$dS_t = S_t (rdt+\theta^2dt+\theta dW_t)$$

implies

$$d\left(\frac{1}{S_t}\right) = \frac{1}{S_t} (-rdt-\theta dW_t)$$

Define $\bar{F}=\bar{F}(t,T):=\frac{F(t,T)}{S_t}$, which implies $\bar{F}(T,T)=\frac{F(T,T)}{S_T}$ and $H_t = \mathbb{E}[\frac{h(L_t)}{S_T}|\mathcal{F}_t]$. $H_t$ is a martingale.

The solution of $Z_t = 1+\int_0^t Z_{s}dXs$ is $\mathcal{E}(-X)_t$ which is just the reciprocal of the savings account, i.e. $\mathcal{E}(-X)_t=\frac{1}{B_t}$. Note that $\mathcal{E}(-X)_t dX_t = -\mathcal{E}(X)_t r dt = -\frac{1}{B_t}r dt = d\frac{1}{B_t} = d\left( \mathcal{E}(-X)_t \right)$.

We get:

$$ 0=H_t + \mathbb{E}\left[ -\bar{F}(t,T) - \int_t^{T}\bar{F}(s-,T)dX_s \big|\mathcal{F}_t\right], $$ which looks very similar to a OU-SDE (except it doesn't run from $0$ to $t$). After looking carefully at the solution of the usual OU-SDE (see Revuz and Yor p. 378 Prop. 2.3), I tried $$\bar{F}(t,T)=\mathbb{E}\left[ \mathcal{E}(-X)_t\left( \frac{H_t}{\mathcal{E}(-X)_T}+\int_t^T\mathcal{E}(-X)^{-1}_s(dH_s-d\langle H,X\rangle_s) \right)|\mathcal{F}_t\right]$$

Since $H_t$ is a martingale and $X_t$ deterministic, we can forget about the integral and obtain: $$ \bar{F}(t,T)=\frac{B_t}{B_T}\mathbb{E}\left[ \frac{h(L_T)}{S_T} |\mathcal{F}_t \right]\Rightarrow F(t,T)=S_t \frac{B_t}{B_T}\mathbb{E}\left[ \frac{h(L_T)}{S_T} |\mathcal{F}_t \right]$$

Plugging in shows that this is indeed a solution. I am not sure of uniqueness though. No solution for random $X_t$ yet, let alone if $\langle X,H\rangle\neq0$.

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