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It is well known, that under the Black-Scholes framework:

$$F\left(t,T\right)=\exp\left(r\left(T-t\right)\right)S\left(t\right),$$

where $S\left(t\right)$ is the spot price of an asset at time $t$, $F\left(t,T\right)$ is the forward price of the same asset at time $t$ with maturity $T$ and $r$ is the constant risk free rate. $S$ and $F$ both follow geometric Brownian motions with constant parameters.

As a consequence of the equation above, I would say that the $\sigma$ parameters are the same for $S$ and $F$ (, i.e. the SDE of $S$ and $F$ have the same $\sigma$ parameters).

However, most of time (in practice and in theory as well) the forward prices with different $T$ maturities are modelled separately with different SDEs. Other words, for each $T$ maturity there is a different SDE of the forward price:

$$dF\left(t,T\right)=\sigma_{F^{T}}F\left(t,T\right)dW_{F^{T}}\left(t\right)$$

under the risk neutral measure, where $\sigma_{F^{T}}$ are different constans for each $T$, and $W_{F^{T}}$ are different Wiener-processes for each $T$. As a result, the $\sigma$ parameter can't be the same for $S$ and $F$, since for all maturities the $\sigma_{F^{T}}$ should match with the $\sigma$ parameter of $S$, even though we assumed $\sigma_{F^{T}}$ parameters can differ.

In short, these two statments above are contradictional to me. How can it be resolved? What do I misunderstand?

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  • $\begingroup$ In the Black-Scholes framework, \begin{align} e^{-rt}S_t&=S_0\,e^{\sigma W_t-\sigma^2t/2}\,. \end{align} Multiply this with the constant $e^{rT}$ to get $F(t,T)$ and check what SDE it solves. It has the same volatility $\sigma$ and the same Brownian motion $W_t\,.$ $\endgroup$
    – Kurt G.
    Dec 16, 2023 at 17:55
  • $\begingroup$ Just make you spot volatility time dependent - then no contradiction. $\endgroup$
    – river_rat
    Dec 16, 2023 at 22:30
  • $\begingroup$ So $\sigma=\sigma_{F^{T_{1}}}=\sigma_{F^{T_{2}}}=\sigma_{F^{T_{3}}}=\ldots$ is true when we use the Black-Scholes assumptions (i.e. when almost every parameter is constant)...but this mentioned equality can fail even for the smallest generalization like it is enough to make the instantaneous vol time dependent and the equation above fails? $\endgroup$
    – Kapes Mate
    Dec 16, 2023 at 23:27
  • $\begingroup$ I recommend to learn a bit of Ito calculus. There is not much required to understand this case. When the volatility is as general as we want and the riskless rate is constant then \begin{align} e^{-rt}S_t&=S_0\,e^{\int_0^t\sigma_s\,dW_s-\frac12\int_0^t\sigma^2_s\,ds}\,. \end{align} Then proceed as in my previous comment to come to the same conclusion. $\endgroup$
    – Kurt G.
    Dec 18, 2023 at 11:42

1 Answer 1

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They are not contradictory, if we choose sigmas reasonably (time-dependent, as suggested in the comments).

If $$ F_t^T = e^{r(T-t)} S_t \; \; {\rm and} \; \; dS_t = rS_tdt + \sigma_t S_t dW_t, $$ with $r$ constant and $\sigma_t$ deterministic function of $t$, then, by Ito Lemma, we get: $$ d F_t^T =\sigma_t F_t^T dW_t.$$ If we focus on a forward maturity structure with only two points, $ t_0< T_1 <T_2$, we could choose $\sigma$ to be, for example, piecewise constant: $$ \sigma_t= \left\{ \begin{array}{ll} \sigma_1 & t_0\leq t \leq T_1\\ \sigma_2& T_1 <t\leq T_2 \end{array} \right. $$ As $F_t^{T_1}$ dies out (zero diffusion coefficient/volatility) once $t>T_1$, its dynamics uses only $\sigma_1$. Similarly, $F_t^{T_2}$ dies out once $t>T_2$ and uses both $\sigma_1$ and $\sigma_2$.

So, using your notations:

$$ d F_t^{T_1} =\sigma^{F^{T_1}}_t F_t^{T_1} dW_t,$$ $$ d F_t^{T_2} =\sigma^{F^{T_2}}_t F_t^{T_2} dW_t,$$ where $$ \sigma^{F^{T_1}}_t := \sigma_t = \sigma_1, $$ for all $t\in [t_0,T_1]$ (which is constant), and $$ \sigma^{F^{T_2}}_t:= \sigma_t = \left\{ \begin{array}{ll} \sigma_1 & t_0\leq t \leq T_1\\ \sigma_2& T_1 <t\leq T_2. \end{array} \right. $$ (which is not constant).

The model you propose also suggests taking control of the Brownian driver of the second forward, which can be done by introducing: $$ W^1_t := W_t \; \; {\rm and} \; \; W_t^2, \; \; {\rm with } \; \; dW^1_tdW^2_t=\rho dt$$ ($\rho$ constant).

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  • $\begingroup$ Thanks for the answer. I think that would be an other good example if the $r$ risk rate would be stochastic. In that case $F\left(t,T\right)=\frac{S\left(t\right)}{p\left(t,T\right)}$, where $p\left(t,T\right)$ is the price of a zero-coupon bond with maturity $T$. In that case, assuming $p\left(t,T\right)$ follows a kind of “generalized” GBM with $\sigma_{p}$ parameter (, for the sake of convenience with the same Wiener process that the spot price has), then... $\endgroup$
    – Kapes Mate
    Dec 19, 2023 at 21:24
  • $\begingroup$ ... according to Itô's formula $d\left(\frac{S\left(t\right)}{p\left(t,T\right)}\right)=\frac{S\left(t\right)}{p\left(t,T\right)}\left(\ldots+\left(\sigma_{S}\left(t\right)-\sigma_{p}\left(t\right)\right)dW\left(t\right)\right)$, so clearly $\sigma_{F}\dot{=}\sigma_{S}-\sigma_{p}\neq\sigma_{S}$. $\endgroup$
    – Kapes Mate
    Dec 19, 2023 at 21:25

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