# Forward price - T-forward martingale

I have a problem figuring out some of the calculations in the book: Fixed Income modelling

In the chapter on forwards the author makes an argument that the forward is a martingale under the T-forward martingale measure.

I know that the forward is given by:

$$F_t^T=\frac{P_t}{B_t^T}\;(1)$$

And that the price of a security under the risk-neutral probability measure, with no payments in the given period is:

$$P_t=E_t^{Q}[e^{-\int\limits_t^Tr_udu}P_T]\;(2)$$

The T-forward martingale measure is: $E_t^{Q^T}$

And in the book we have:

$$P_t=B_t^TE_t^{Q^T}[P_T]\;(3)$$

First question: What is the difference between (2) and (3)? $B_t^T =e^{-\int\limits_t^Tr_udu}$, so how do they differ?

Next he says that with $B_t^T$ as a numeraire we have that:

$$\frac{0}{B_t^T}=E_t^{Q^T}[\frac{P_T-F_t^T}{B_T^T}];(4)$$

How does he get that? More specifically. Why is $B_T^T$ in the equation as that is equal to 1.

From (3) i have: $P_t=B_t^TE_t^{Q^T}[P_T]$

Subtracting $P_t$ on both sides: $0=B_t^TE_t^{Q^T}[P_T]-P_t$

Dividing by $B_t^T$ and using (1) I get:

$\frac{0}{B_t^T}=E_t^{Q^T}[P_T]-F_t^T$

So where does the $B_T^T$ come from in (4)? It is equal to 1, so I know I can always divide by it. But why?

• From (1), the forward price is a ratio of asset prices, with the price of a T-zero coupon bond in the denominator. Therefore the forward price is s martingale under the T-forward measure. – dm63 Feb 3 '17 at 11:20
• Think about stochastic interest rates. $B_t^T$ is a number while $e^{-\int\limits_t^Tr_udu}$ is a random variable. The former is the conditionnel expectation under the risk-neutral measure $Q$ of the latter. – Quantuple Feb 3 '17 at 11:37
• So there really isn't any difference between $E_t^{Q^T}$ and $E_t^Q$? @Quantuple – Anders Feb 3 '17 at 12:50
• @Anders: $B_t^T = \Bbb{E}_t^Q \left[e^{-\int\limits_t^Tr_udu}\right]$, such that there is a difference between the measures $Q$ and $Q_T$ when interest rates are stochastic. $Q$ uses the $t$-value of the money market account as numéraire, $Q_T$ uses the price of a $T$-zero coupon bond. – Quantuple Feb 3 '17 at 14:03
• thanks @Quantuple. That cleared a lot for me. I still don't get equation 4 though – Anders Feb 3 '17 at 17:10