Is the money market account (MMA) numeraire and the forward measure equivalent?

Question

Suppose we have a risk-neutral measure $\tilde{\mathbb{P}}$. The money market account is given as $M(t) = e^{\int^t_0 R(s) ds}$, while the price of the zero-coupon bond at time $t$ that matures at $T$ is denoted $B(t,T)$.

So, the forward measure is defined to be the measure with $B(t,T)$ taken as the numeraire. However, I am curious if taking $M(t)$ will also make the measure into a forward measure. If this is not true in general, does it work when the interest rate is constant as $R(t) = r$? This would imply that $B(t,T) = e^{r(T-t)}$, and $B(0,T) = \frac{1}{M(T)}$ and $B(T,T) = \frac{1}{M(0)}$, which seems to imply somewhat of a connection between the two measures just by looking at the Radon-Nikodym derivative, $\mathbb{Z}$.

Also, I have an additional question about the usefulness of the forward measure. It seems that forward measures are useful in options pricing because we can take the discount out for the risk-neutral pricing formula so that $V(t) = D(t) \tilde{\mathbb{E}}^F[V(T) | {\cal{F}}(t)]$. But are there any other advantages of using the forward measure?

Answer 1

Your questions are nicely addressed in this short paper by Fabrice Rouah: The T-forward measure

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More specifically, using your notations and noting that $B(T,T)=1$ by definition, the change of measure between the $T$-forward ($\mathbb{Q}^B$) and risk-neutral ($\mathbb{Q}^M$) measures is characterised by the following Radon-Nikodym derivative: $$\left. \frac{d \mathbb{Q}^B}{d \mathbb{Q}^M } \right\vert_{\mathcal{F}_t} = \frac{M(t)B(T,T)}{M(T)B(t,T)} = \frac{M_t}{M_T}\frac{1}{B(t,T)} $$

Yet, by construction of the martingale measure $\mathbb{Q}^M$, the following relationship for the price of a zero-coupon bond prevails \begin{align} \frac{B(t,T)}{M_t} &= E^{\mathbb{Q}^M}_t\left[ \frac{B(T,T)}{M_T} \right] \\ B(t,T) &= E^{\mathbb{Q}^M}_t\left[ \frac{B(T,T) M_t}{M_T} \right] \\ B(t,T) &= E^{\mathbb{Q}^M}_t\left[ \frac{M_t}{M_T} \right] \end{align} where I have used the notation $E_t[.]$ to represent $E[.\vert\mathcal{F}_t]$.

Plugging the above result in the expression of the Radon-Nikodym derivative gives: $$ \left. \frac{d \mathbb{Q}^B}{d \mathbb{Q}^M } \right\vert_{\mathcal{F}_t} = \frac{M_t}{M_T}\frac{1}{B(t,T)} = \frac{M_t}{M_T}\frac{1}{E^{\mathbb{Q}^M}_t\left[ M_t/M_T \right]} = \frac{1/M_T}{E^{\mathbb{Q}^M}_t\left[ 1/M_T \right]}$$

When interest rates are deterministic, then $$ E^{\mathbb{Q}^M}_t\left[ \frac{1}{M_T} \right] = \frac{1}{M_T} $$ and $$ \left. \frac{d \mathbb{Q}^B}{d \mathbb{Q}^M } \right\vert_{\mathcal{F}_t} = 1 $$ such that the measures $\mathbb{Q}^B$ and $\mathbb{Q}^M$ are perfectly equivalent.

With stochastic interest rates, this is not true any more.

Answer 2

Maybe a minor detail, but it's not the discount factor that goes in front, but the bond price, so $$V(t)=P(t,T) \, E^{F}[V(T)|\mathcal{F}(t)].$$ $P(T,T) = 1$ so this can be written as $$V(t)=P(t,T) \, E^{F}[V(T) / P(T,T)|\mathcal{F}(t)].$$

Compare this with $$V(t) = B(t) \, E[V(T)/B(T)|\mathcal{F}(t)]$$ in the measure of the cash account.

If you have stochastic interest rates, $B(T)$ is a stochastic quantity. In a Monte Carlo simulation for example, you only have to simulate $V(T)$ in the case of the forward measure, but $V(T)$ and $B(T)$ in the cash measure. So replacing $B(T)$ with $P(T,T) = 1$ can be a very attractive thing.