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I recently went through some commodities forward curve modeling documentations, where a diffusion model for the forward price $F(t,T)$ was modeled as a driftless diffusion process (as a function of t with T fixed). The document did not mention whether this model is under risk-neutral measure or real-world measure. The model was estimated using historical data assuming trendless. It was later also used for derivative pricing, which is supposedly under the risk neutral measure. For privacy purposes I cannot reveal the source of this document, but just wonder if it is the case that for commodities, these two measures are the same? Is the forward price expected to not change over time even under the real-world measure? If so, what is the argument? Risk premium equal to zero in the commodities world?

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Futures prices are drift less under risk neutral measure. In commodities market, it is often Futures. They need to estimate volatility in their model. Since volatilities are not affected by change of probability, you can estimate under real word measure. So what you describe seems correct.

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To satisfy commentators.

In the continuous time semi martingale framework of Girsanov theorem, a change of probability measure (equivalent) affects only the finite variation part. So of course, if you speak about stochastic volatility models, then the sentence is not completely true. Volatilities have to be understood as diffusion part (the $\sigma_t$ in front of $dW_t$)

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    $\begingroup$ I do not completely agree. Volatility will be affected by a change of measure if it is modelled as a separate stochastic process with its own driving Brownian motion. Think of Heston for instance, the volatility dynamics under $\Bbb{P}$ and $\Bbb{Q}$ will not be the same (the long term mean and mean reversion speed change). $\endgroup$
    – Quantuple
    Commented Nov 24, 2016 at 9:19
  • $\begingroup$ @Quantuple Yes. In the Heston model, we have $$dS_t=rS_tdS_t+\sqrt{v_t}S_tdW_1^{\mathbb{Q}}(t)\\ dv_t=\kappa^*(\theta^*-v_t)+\sigma\sqrt{v_t}dW_2^{\mathbb{Q}}(t)$$ where $\kappa^*=\kappa+\lambda$ and $\theta^*=\frac{\kappa\theta}{\kappa+\lambda}$ where $\lambda$ is the Volatility Risk Primuim. $\endgroup$
    – user16651
    Commented Nov 24, 2016 at 10:43
  • $\begingroup$ If $\lambda=0$ then $\kappa^*=\kappa$ and $\theta^*=\theta$ $\endgroup$
    – user16651
    Commented Nov 24, 2016 at 10:47
  • $\begingroup$ Exactly @BehrouzMaleki :) So volatility is indeed affected by the change of measure contrary to the standard BS case. $\endgroup$
    – Quantuple
    Commented Nov 24, 2016 at 11:52
  • $\begingroup$ Ok. to be fully clear, in a SDE only the drift term is affected by the change of measure. $\endgroup$ Commented Nov 24, 2016 at 13:19
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Definitions

For fixed $T$ and moving $t \leq T$ then by definition $\color{blue}{(*)}$, forward prices $F(t,T)$ and future prices $\text{Fut}(t,T)$ are both conditional expectations. However, these expectations are not taken under the same probability measure. More specifically: $$ F(t,T) = \Bbb{E}^{\Bbb{Q}^T}\left[ \left. S_T \right\vert \mathcal{F}_t \right]$$ $$ \text{Fut}(t,T) = \Bbb{E}^{\Bbb{Q}}\left[ \left. S_T \right\vert \mathcal{F}_t \right] $$ where

  • $\Bbb{Q}$ the risk-neutral measure: measure under which the $t$-value of all self-financing portfolio emerges as a martingale when expressed relative to the money market account numéraire $B_t = \exp(\int_0^t r_s ds)$
  • $\Bbb{Q}^T$ denotes the $T$-forward measure: measure under which the $t$-value of all self-financing portfolio emerges as a martingale when expressed with respect to the zero-coupon $T$-bond numéraire $P(t,T) = \Bbb{E}^\Bbb{Q}_t\left[B_t B_T^{-1}\right]$.

From the above definitions we have that:

  • When interest rates are deterministic, both measures coincide. This can be seen by writing the Radon-Nikodym derivative of the change of measure: $$ \left. \frac{d\Bbb{Q}}{d\Bbb{Q}^T} \right\vert_{\mathcal{F}_t} = \frac{P(0,T) B_t}{B_0 P(t,T)} $$
  • Being simple conditional expectations, forward and future prices are martingales under their respective measures. This is a direct consequence of the tower property of conditional expectations. Indeed, looking at the future price process without loss of generality, for $s < t$ one can always write: $$\Bbb{E}^{\Bbb{Q}^T}\left[ \left. \text{Fut}(t,T) \right\vert \mathcal{F}_s \right] = \Bbb{E}^{\Bbb{Q}^T}\left[ \Bbb{E}^{\Bbb{Q}^T}\left[ \left. \left. S_T \right\vert \mathcal{F}_t \right] \right\vert \mathcal{F}_s \right] = \Bbb{E}^{\Bbb{Q}^T}\left[ \left. S_T \right\vert \mathcal{F}_s \right] = \text{Fut}(s,T) $$

$\color{blue}{(*)}$ e.g. the forward price is defined such that the value of a forward contract is zero at inception. Denoting $t$ the inception date, the forward price is thus the strike $K$ such that the expected discounted cash flows of the contract $\Bbb{E}^{\Bbb{Q}}_t \left[ B_t B_T^{-1} (S_T-K) \right] = 0$, which is equivalent to claiming that $F(t,T) = \Bbb{E}_t^{\Bbb{Q}^T} [S_T]$ with $\Bbb{Q}^T$ the measure associated to the zero coupon $T$-bond numéraire $P(t,T)$ since $P(t,T)$ is $\mathcal{F}_t$-measurable. A similar reasoning can be used for futures (but the daily settlement mechanism makes it a bit trickier to write down).


Answer

To simplify things assume deterministic interest rates (so that $\Bbb{Q} = \Bbb{Q}^T$) and that we only manipulate adapted, continuous paths processes that verify the usual conditions.

From the above definitions, forward/future prices are martingales under some measure $\Bbb{Q}$ equivalent to the real world measure $\Bbb{P}$. By the martingale representation theorem, this means that they are driftless. Indeed we have that: $$ F(t,T) = \Bbb{E}^\Bbb{Q}_t [ S_T ] \iff dF(t,T) = \sigma_t dW_t^\Bbb{Q} $$

By Girsanov theorem (or Abstract Bayes), this also means that they are not driftless under the equivalent real-world measure $\Bbb{P}$ in general. Indeed, $$ F(t,T) = \Bbb{E}^\Bbb{P}_t [ S_T ] = \Bbb{E}^\Bbb{Q}_t \left[ S_T \frac{Z_T}{Z_t} \right] \iff dF(t,T) = \mu dt + \sigma_t dW_t^\Bbb{P} $$

where $\mu$ is given by the differential of the quadratic covariation between $W_t^\Bbb{Q}$ and the stochastic logarithm of the change of measure process $\frac{Z_T}{Z_t}$, where $Z_t= \left. \frac{d\Bbb{P}}{d\Bbb{Q}}\right\vert_{\mathcal{F}_t}$ $$ \mu = d\langle W_t^\Bbb{Q}, \mathcal{L}(Z_T/Z_t) \rangle_t $$

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    $\begingroup$ very good answer! $\endgroup$
    – Richi Wa
    Commented Nov 24, 2016 at 10:43
  • $\begingroup$ While I like this answer, I think it should be stressed that this answer is not entirely general: it assumes that asset prices are diffusions while in all likelihood they have discontinuous (in space if not time) paths. $\endgroup$
    – user9403
    Commented Nov 24, 2016 at 16:46
  • $\begingroup$ @user9403. Assume today is $t=0$ and $T=1Y$, further assume that the spot price jumps e.g. a large dividend is paid by the stock at $t^{ex}=0.5$. Do you mean to say that $F(t,T)$ will jump over $[0,T]$ (which is not true)? Or are you thinking of something else? $\endgroup$
    – Quantuple
    Commented Nov 24, 2016 at 17:16
  • $\begingroup$ Thanks for the detailed calculations., especially about the difference between forwards and futures. I suppose the model I was looking at which is a local volatility model is under the measure $Q$. Why can we estimate the local volatility using the historical data? Is it because under this model, the measure change does not affect the local volatility, at least by assumption? $\endgroup$
    – user138668
    Commented Nov 24, 2016 at 23:30
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    $\begingroup$ @Quantuple My point was there are continuous time stochastic processes that are not diffusions (levy processes). Diffusions are continuous in both time and space, but more generic levy processes are only continuous in time (they can "jump" in space). $\endgroup$
    – user9403
    Commented Nov 25, 2016 at 13:53
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The answer given by MJ73550 already covers most if the points in my opinion.

I would put it like this:

Concerning the drift: if the cost-of-carry relationship is used in your model then this is the correct drift to use for the spot price to derive the price any derivatives (forwards, futures, options) - this is the risk neutral drift. This has nothing to do with the real world drift (which you can only guess/model for the future or observe from traded prices for the past).

If you look at forwards then you imply the drift of the spot. Sometimes this is different to the cost-of-carry relationship (in commodity markets it might also be difficult to determine the convenience yield).

Thus the spot has some drift and it is fixed either by the forwards traded or cost-of-carry.

So far having modelled the drift of the spot in the risk neutral world we recall that we do not use it to guess the future but to make the prices of traded instruments match.

For a forward or futures contract we don't need a risk neutral drift to match traded prices. The futures price already is the correct price for a linear derivative of the spot at some point in time in the future and that's it. But it still has volatility.

Then we have two approaches:

  • if we want to price e.g. options on these futures then we have to use implied volatility (derived from other options on the same commodity). Then again prices will fit together. This is risk neutral.

  • if we want to estimate the risk that we have from this futures position then we talk about the real world. Then we could look at historical volatility.

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  • $\begingroup$ Thank you for the clarification. So the cost-of-carry drift only affects spot price but not future price because no cost-of-carry is incurred by holding a future contract? Concerning the implied volatility: if few option data is available ( which is often the case in some commodities), is it customary to use the realised (estimated from historical data) volatility for pricing purposes? $\endgroup$
    – user138668
    Commented Nov 24, 2016 at 23:28
  • $\begingroup$ concerning the drift: the forward is priced using its cost of carry and this is the drift of the spot. @implied vol: when you price options you look into the future, historical vol is backwards looking. I would not use that. $\endgroup$
    – Richi Wa
    Commented Nov 25, 2016 at 8:14

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