# T-Forward Price on risk-neutral measure

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

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it seems to me that risk neutral pricing implies that $F(t, T) = E^\star\left[\frac{S^1_T}{S^0_T}|\mathscr{F}_t\right]$ perhaps this makes things a little more clear –  user25064 Feb 3 at 14:09
i'm not sur about what you're saying @user25064. Because under the risk-neutral measure $P^{*}$, the discounted risky asset is an martingale. That mean the left hand side in your equality is not the Forward price, but has to be the discounted price of the risky asset at time t, i.e $\frac{S^1_t}{S^0_t}$ instead of $F(t,T)$ –  ctNGUYEN Feb 3 at 17:22
You are connect, in fact they are the same. As you say in your post, "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" If I read this correctly, this is a derivative security with payoff $S^1_T$ at time $T \geq t$ in this case the payoff of the derivative is the identity function $f(x) = x$ assuming for the moment that the asset pays no dividends, what I wrote above is the fair value of such a contract. –  user25064 Feb 3 at 18:19
Thanks user25046. Well, i now understand the issue. user25046, you are talking about a problem of giving a claim $h_{T}$ at time $T$, what is its fair price at time t. The forward price is not exactly the same. That is the price decided for $h_{T}$, agreed at time $t$, but paid at time $T$. It is this subtile that make me confused. For people who interest to understand the question, i've found more explain in the book of Marek Musiela and Marek Rutkowski - "Martingale Methods in Financial Modelling", sec 9.6.1 p.373. –  ctNGUYEN Feb 3 at 20:45
AH I see the difference, the price is paid at time $T$! That is the subtle difference! –  user25064 Feb 3 at 22:54

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.

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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 $t$. $$E_{P^{*}}(\tilde{S}_{t+\Delta}|I_{t}) = E_{P^{*}}(\frac{S_{t+\Delta} - F(t+\Delta,T)}{S_{0}})\ldots(1)$$ If we assume that the return on stock price is a weiner process, then using the linearity property of expectations we have that, $$E_{P^{*}}(S_{t+\Delta}/S_{0}) = S_{t}.$$ 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, $$E_{P^{*}}(\tilde{S}_{t+\Delta}|I_{t}) = \tilde{S}_{t}.$$ 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.

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