# Stochastic Interest rate spot forward relationship

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?

edit

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?

• You may want to have a look of this question. – Gordon Mar 27 '17 at 18:23

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*}

• 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? – Calypso Mar 28 '17 at 6:46
• I don't assume deternistic interest rate. $P (t,T)$ should be the conditioanl expectation. $B_t$ is correct. – Gordon Mar 28 '17 at 12:48
• 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 – Calypso Mar 28 '17 at 20:13
• The independence assumption is not needed. – Gordon Mar 30 '17 at 3:08