T-Forward Price on risk-neutral measure

Question

i have and question concerning the T-forward price definition on the Robert J.Elliot's book : Mathematics of Financial Markets. On his chapter 9, definition 9.1.3 p.249. He give the formula without explaining how can it possible. I've tried to understand why but without success, so i post the question here to request your help.

Giving two asset $S^1$ and $S^0$ which are risky and riskless asset, $P^{*}$ is the risk-neutral probability on what the discounted risky asset is a martingale (after Girsanov transform).

He defined T-Forward price $F(t,T)$ as the price of the risky asset agreed at time $t\leq T$ that will be paid for $S^1$ at time $T$. Then he says that such a price satisfied that the claim $S^1_T - F(t,T)$ is a martingale (under risk-neutral probability), more strongly, a zero martingale. Then he give the formula
$$ 0=E^{*}\left(\frac{S^1_T - F(t,T)}{S^0_T}\bigg|\mathscr{F}_t\right) $$ This argument confused me. After what i learned, the claim which is zero has to be $S^1_T - F(T,T)$, we can have it because the forward price converge to the spot price at maturity time. So the formula has to be
$$ 0=E^{*}\left(\frac{S^1_T - F(T,T)}{S^0_T}\bigg|\mathscr{F}_t\right) $$ in my sens. But if i do what i thought, i can not manipulate the formula in order to have the final result as he mentioned so far in the book $$ F(t,T) = \frac{S^1_t}{B(t,T)} $$
where $B(t,T)$ is the zero coupon bond.

This point is important for me because it help to understand all theory of numeraire change, and the T-Forward price for another claim. Can anyone have some suggest please? Thanks

Answer 1

For the buyer of a forward contract the payoff is $S_T - K$ at time $T$ since at this date he pays $K$ and gets the underlying in exchange. Consider the following strategy: buy the stock $S$ and sell $K$ zero-coupon bonds with maturity $T$. At any time $t$, your portfolio's value is $$ \Pi_t = S_t - KB(t,T) $$ In particular at time $T$, it is $S_T - KB(T,T) = S_T - K$ the price of our forward contract so by abscence of arbitrage the forward contract must have the same value as our portfolio at each $t \leq T$ (otherwise you could buy the cheaper one, sell the other, invest the difference at the risk free rate and make almost surely a profit at time $T$).
$$ Forward_t = S_t - KB(t,T) $$ By definition the $T$-forward price $F_t^T$ is the "fair strike" $K$ set at $t$ so that the value at $t$ of the forward contract is zero. Clearly we must set
$$ K = F_t^T = \frac{S_t}{B(t,T)} $$

Let's take a look at the martingale approach. Just like for any contingent claim we have the martingale property $$ \frac{Forward_t}{S^0_t} = E^*[ \frac{Forward_T}{S^0_T} | \mathcal{F}_t ] = E^*[ \frac{S_T - K}{S^0_T} | \mathcal{F}_t ] $$ which is another way of saying we have the pricing formula $$ Forward_t = S^0_t E^*[ \frac{S_T - K}{S^0_T} | \mathcal{F}_t ] = E^*[e^{-\int_t^T r_s ds}(S_T -K) | \mathcal{F}_t ] $$ Separating the $S$ and $K$ parts we get $$ Forward_t = S^0_t E^*[ \frac{S_T}{S^0_T} | \mathcal{F}_t ] - KS^0_t E^*[ \frac{1}{S^0_T} | \mathcal{F}_t ] = S_t - K B(t,T) $$ The last equality follows

• for the first term from the fact that $\frac{S}{S^0}$ is a martingale: $E^*[ \frac{S_T}{S^0_T} | \mathcal{F}_t ] = \frac{S_t}{S^0_t} $ (we assume $S$ pays no dividend).
• for the second term from the definition of $B(t,T)$ as the price of a contract that pays 1 at time $T$.

And once again we find that $T$-forward price of $S$ at time $t$ is $$ F_t^T = \frac{S_t}{B(t,T)} $$

Note that it is not true that the formula "has to be" $$ 0=E^{*}\left(\frac{S^1_T - F(T,T)}{S^0_T}\bigg|\mathscr{F}_t\right) $$ Just because its (conditional) expectation has to be zero doesn't mean the random variable has to be zero! Obviously the $T$-forward price at time $T$ is $F(T,T) = S_T$ but at $t < T$ the forward price $F_t^T$ is higher than the spot price $S_t$ and it can be higher or lower than what $S_T$ will eventually be.

Hope that helps.

Answer 2

First lets analyse the claim that $\frac{(S_t - F(t,T)}{S_{0}}$ is a martingale under a given risk neutral measure $P^{*}$. Recall that the crucial property of a martingale is that at some point in time $t$, a process $\tilde{S}_{t}$ is a martingale iff for some time $t+\Delta$, the expected value of $\tilde{S}_{t+\Delta}$ is $S_t$.

So lets start, assume we have a filteration denoted by $I_t$ -- here intuitively $I_t$ corresponds to information that we have at time $t$ in our probability space.

Let $\tilde{S}_{t} = \frac{(S_{t+\Delta} - F(t+\Delta,T)}{S_{0}}$. Now lets take the expected value of this process with respect to the information we have at time $t$. \begin{equation} E_{P^{*}}(\tilde{S}_{t+\Delta}|I_{t}) = E_{P^{*}}(\frac{S_{t+\Delta} - F(t+\Delta,T)}{S_{0}})\ldots(1) \end{equation} If we assume that the return on stock price is a weiner process, then using the linearity property of expectations we have that, \begin{equation} E_{P^{*}}(S_{t+\Delta}/S_{0}) = S_{t}. \end{equation} Similarly, under the risk neutral probability measure we have that $F(t+\Delta ,T) = F(t,T)$ -- this is by definition of the risk neutral forward price for time $[t,T]$. If we substitute this into the equation (1) we get that, \begin{equation} E_{P^{*}}(\tilde{S}_{t+\Delta}|I_{t}) = \tilde{S}_{t}. \end{equation} This proves the desired result -- the process $\tilde{S}_t$ is a martingale.

To see what this really means, lets analyse what $\tilde{S}_{t}$ comprises of. In particular $\tilde{S}_{t}$ comprises of stock and bonds. Using your notation here, we are long a single stock given by $S_t$ and short a bond with yield $F(t,T)$. If this portfolio replicates the payout of the derivative security, as we say it does, then the expected price of the security at time $t+\Delta$ is the price of the portfolio at time $t$ -- this is because of the martingale property proved above. In fact more is true. Since we are pricing this under a risk neutral measure, we can also say that this expected price is in fact the arbitrage free price of the portfolio, and hence the arbitrage price of the underlying security.