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For a stock that pays no dividends, and has constant interest rates, I know that the relationship can be described with:

$F(t) = s(t). e^{r(T-t)}$

F(t) = Forward price, S(t) = Stock price, r = risk free rate, T = maturity date, t = time now

How does the equation change, when rates are stochastic?

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From another textbook, it appears the relationship can be modeled as:

$ F(t) = s(t) / P(t) $

where $ P(t) = e^{-\int_{t}^{T} r(s) ds} $

does this look right?

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  • $\begingroup$ You may want to have a look of this question. $\endgroup$
    – Gordon
    Mar 27, 2017 at 18:23

1 Answer 1

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The forward price $K$, determined at time $t$, is the amount such that the payoff at time $T$ is $S_T-K$, while the value at time $t$ is zero. That is, \begin{align*} B_t E\left(\frac{S_T-K}{B_T}\mid \mathcal{F}_t \right)= 0, \end{align*} Where $E$ is the risk-neutral expectation operator. Then, \begin{align*} K&=\frac{E\left(\frac{S_T}{B_T}\mid \mathcal{F}_t \right)}{E\left(\frac{1}{B_T}\mid \mathcal{F}_t \right)}\\ &=\frac{E\left(\frac{S_T}{B_T}\mid \mathcal{F}_t \right)}{\frac{1}{B_t}E\left(\frac{B_t}{B_T}\mid \mathcal{F}_t \right)}\\ &=\frac {\frac{ S_t}{ B_t} }{\frac{1}{B_t}P (t,T)}\\ &=\frac{S_t}{P(t,T)}, \end{align*} where \begin{align*} P (t,T) &= E\left(\frac{B_t}{ B_T}\mid \mathcal {F}_t \right)\\ &=E\left(e^{-\int_t^T r_s ds}\mid \mathcal {F}_t \right) \end{align*} is the price at time $t$ of a zero-coupon bond with maturity $T$ and unit face value.

Alternatively, at time $t$,

  • enter into a forward contract with forward price $K$, which has zero cost at time $t$,

  • short one share with income $S_t$, and

  • long $\frac{S_t}{P (t,T)}$ units of zero-coupon bond with maturity $T$.

The net cost at time $t$ is zero. At maturity $T$,

  • the forward contract has value $S_T-K$,

  • the short position of one share has value $-S_T$, and

  • the zero-coupon bond has value $\frac{S_t}{P (t,T)}$.

Assuming arbitrage free, the value at time $T$ is then \begin{align*} S_T-K - S_T + \frac{S_t}{P (t,T)} = \frac{S_t}{P (t,T)}-K =0, \end{align*} that is, \begin{align*} K=\frac{S_t}{P (t,T)}. \end{align*}

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  • $\begingroup$ Thanks Gordon, that is a great answer. Please can you confirm that $P(t,T)$ is the price at time t of a zero coupon bond that pays 1 dollar at time T, i.e $$P(t,T) = e^{ - \int_{t}^{T} r(s) ds}$$ and $B_t$ is the time value of the money market account i.e $$B_t = e^{ \int_{0}^{t} r(s) ds}$$ Finally, in this derivation, are you assuming interest rates to be deterministic? $\endgroup$
    – user3387245
    Mar 28, 2017 at 6:46
  • $\begingroup$ I don't assume deternistic interest rate. $P (t,T)$ should be the conditioanl expectation. $B_t$ is correct. $\endgroup$
    – Gordon
    Mar 28, 2017 at 12:48
  • $\begingroup$ Thanks Gordon, to summarise then: $$ $$ $P(t,T)$ is the expected (using risk neutral measure) price at time t of a zero coupon bond that pays 1 dollar at time T, i.e $$ P(t,T) = E_Q [ e^{ - \int_{t}^{T} r(s) ds} ] $$ $B_t$ is the time value of the money market account i.e $$ B_t = e^{ \int_{0}^{t} r(s) ds} $$ Assumptions:$$ $$ 1. Stock price and the interest rate are independent. 2. The market is frictionless, arbitrage free, infinite liquidity and complete $\endgroup$
    – user3387245
    Mar 28, 2017 at 20:13
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    $\begingroup$ The independence assumption is not needed. $\endgroup$
    – Gordon
    Mar 30, 2017 at 3:08

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