Timeline for why futures contract has no value
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 14, 2021 at 6:53 | comment | added | Quantuple | Entering a future is done at zero cost indeed, so there is no "upfront" premium $V(t,T)$ to be paid. | |
| Apr 13, 2021 at 17:53 | comment | added | Gabriele Pompa | @Quantuple I'm trying to following your reasoning and I get (using linearity of expectation, tower-law and ${\mathbb Q}$-martingality of futures price) that $V(t,T) = 0$ provided that the discount factor between two subsequent trading dates is neglected, i.e. $D(t_{i-1}, t_{i} ) = e^{ -\int^{t_{i}}_{t_{i-1}} r_s ds } \approx 1$. Can you confirm or comment please? | |
| Mar 13, 2017 at 13:36 | comment | added | A.Oreo | @ChrisTaylor Quantuple, why the change of portfolio of future is $dF$ i,e the change of the future price | |
| Mar 13, 2017 at 13:23 | comment | added | A.Oreo | @Quantuple yeah, I mean the method, in forward, we first solve value of forward $V(t,T),$ then set $V(t,T) = 0,$ we obtain the forward price $\bar{S};$ But in future, why author directly set $F = 0,$ then the solution of PDE is exactly the future price. I think it should be like the forward, first solve value of future, then set as zero to obtain the future price. | |
| Mar 13, 2017 at 12:19 | comment | added | Quantuple | @A. Oreo in your reference $\bar{S}=F(t_0,T)$ the forward price at $t_0$ for delivey at $T$. The value of the forward contract signed at $t_0$ is then $V(t_0,S_0) = e^{-r(T-t_0)} \Bbb{E}^Q \left[ S_T - F(t_0,T) \mid \mathcal{F}_{t_0} \right] = 0$. | |
| Mar 13, 2017 at 11:33 | comment | added | A.Oreo | @Quantuple why for the forward contract, $F$ is not zero in second picture, I think they should be similar | |
| Mar 13, 2017 at 11:24 | comment | added | Quantuple | @ChrisTaylor - Never seen it either I must say. As I said I got the point you were trying to make... I guess I was just being pedantic :) | |
| Mar 13, 2017 at 11:20 | comment | added | Chris Taylor | @Quantuple Yes, this is a valid point! It takes the financing into account very explicitly. In practice though, I think that most people use the first formula in my answer, and then have a correction between values of $F$ derived from futures and forward contracts where necessary (which is really only for futures on interest rates). Perhaps some desks now do something more advanced to take the financing into account, but I haven't personally seen it. | |
| Mar 13, 2017 at 11:19 | comment | added | Quantuple | $F(t,T)$ is zero neither for a futures nor a forward. However the current values $V(t,T)$ of the contract should you enter them at $t$ are. See also here quant.stackexchange.com/questions/31162/…. | |
| Mar 13, 2017 at 11:17 | comment | added | A.Oreo | Pls see the update, why for the forward contract, $F$ is not zero, I think they should be similar. | |
| Mar 13, 2017 at 11:13 | comment | added | Quantuple | Although I of course agree with the point you are trying to make, I think that the equations you propose are in fact better suited for a "forward" contract than a "future" contract. For a future you would have: $$V(t,T) = \Bbb{E}^{\Bbb{Q}} \left[ \sum_{i=1}^N e^{-r(t_i-t)} F(t_{i-1},T)-F(t_i,T) \mid \mathcal{F}_t \right] $$ for each open business day $i=1,...,N$ between now ($t$) and the expiry ($T$) with $F(t,T)$ figuring the future price where indeed $F(t,T) = \Bbb{E}^{\Bbb{Q}^T}[S_T \mid \mathcal{F}_t]$, such that $F(t,T)$ is a martingale and $V(t,T)=0$ | |
| Mar 13, 2017 at 11:03 | history | answered | Chris Taylor | CC BY-SA 3.0 |