Two different ways of pricing that leads to two answers

Question

This question might appear trivial to many (considering the questions on this site), but I think it reflects something fundamental that I am missing.

To keep things simple, assume everyone is risk-neutral and there is no inflation etc, so that prices are determined using expected values.

Let us be in year 0. Consider an asset that pays \$1 in year 1 and \$1 in year 2. Let $r_{01}$ and $r_{02}$ denote the (annually compounded) 1-year spot rate and 2-year spot rate respectively. The standard textbook way to price this asset is:

$$P_0 = \frac{1}{1+r_{01}} + \frac{1}{(1+r_{02})^2}$$

Here is an alternative approach that seems reasonable. We have $P_0 = \frac{1}{1+r_{01}} + \frac{E[P_1]}{1+r_{01}}$ where $P_1$ is the expected price the asset will fetch in year 1. At time 0, $P_1$ is still a random variable and it is calculated by $P_1 = \frac{1}{1+r_{12}}$ where $r_{12}$ is the one-year spot rate when we are in year 1 (hence this is a random variable at year 0). In this case,

$$P_0 = \frac{1}{1+r_{01}} + \frac{E[P_1]}{1+r_{01}} = \frac{1}{1+r_{01}} + E\left(\frac{1}{1+r_{12}}\right)\frac{1}{1+r_{01}}$$

If these two pricing methods are to be equal, then one needs

$$ \frac{1}{(1+r_{02})^2} = E\left(\frac{1}{1+r_{12}}\right)\frac{1}{1+r_{01}}$$

It does not appear to me that this must be so. In fact, I think it is the case that $(1+r_{01})E(1+r_{12})=(1+r_{02})^2$ [by considering the expected amount \$1 under a two-year strip or rollover one-year strips will earn; if LHS>RHS, then borrow \$1 cash using $(1+r_{02})^2$ units of 2-year strips and lend out all that cash on $1+r_{01}$ units of 1-year strips and lent out all earnings again at end of year 1 on $(1+r_{01})(1+r_{12})$ units of 1-year strip to make profit in year 2 in expectation]. Since $E(1/X)\neq 1/E(X)$ in general, the two pricing methods cannot lead to equal outcomes. How do I resolve this contradiction?

Answer 1

$$\frac{1}{(1+r_{02})^2} = E\left(\frac{1}{1+r_{12}}\right)\frac{1}{1+r_{01}}$$

Indeed, in the pricing measure, the distribution of $r_{12}$ has to be such that this relation holds.

If you look at drift derivations for the LIBOR market model, a lot of work goes into making this sort of equation hold.

Answer 2

There is no conflict here. In the identity, \begin{align*} \frac{1}{(1+r_{02})^2} = E\left(\frac{1}{1+r_{12}}\right)\frac{1}{1+r_{01}}, \end{align*} the expectation is under the year-1 forward measure. However, in the identity \begin{align*} (1+r_{01})E(1+r_{12})=(1+r_{02})^2, \end{align*} the expectation is under the year-2 forward measure.

For illustration, let $P(t, u)$ be the price at time $t$ of a zero-coupon bond with maturity $u$ and unit face value. Moreover, let $B_t$ be the money market account (or deposit account) value at time $t$. For notations, let $T_1=1$ and $T_2=2$. Then \begin{align*} r_{12} &\triangleq L(T_1; T_1, T_2)\\ &=\frac{1}{T_2-T_1}\left(\frac{P(T_1, T_1)}{P(T_1, T_2)}-1\right)\\ &=\frac{1}{P(T_1, T_2)}-1. \end{align*} We also note that $ P(0, T_1) = \frac{1}{1+r_{01}}, $ and $ P(0, T_2) = \frac{1}{(1+r_{02})^2}. $ Let $E$, $E_1$, and $E_2$ be respectively the expectation operators under the risk-neutral measure $P$, the year-1 forward measure $P_1$ and the year-2 forward measure $P_2$.

It is well known that the process $\{L(t; T_1, T_2) \mid 0\leq t \leq T_1 \}$ is a martingale under the year-2 forward measure. Then \begin{align*} E_2(1+r_{12}) &= E_2(1+L(T_1; T_1, T_2))\\ &= 1+ L(0; T_1, T_2)\\ &= \frac{P(0, T_1)}{P(0, T_2)}\\ &= \frac{(1+r_{02})^2}{1+r_{01}}. \end{align*} That is, \begin{align*} (1+r_{01})E_2(1+r_{12})=(1+r_{02})^2. \end{align*}

On the other hand, note that, for $0 \leq t \leq T_1$, \begin{align*} \frac{dP}{dP_1}\big|_t = \frac{B_t P(0, T_1)}{P(t, T_1)}. \end{align*} Then, \begin{align*} \frac{1}{(1+r_{02})^2} &= P(0, T_2)\\ &= E\left(\frac{1}{B_{T_2}}\right)\\ &= E\left(\frac{1}{B_{T_1}}E\left(\frac{B_{T_1}}{B_{T_2}} \mid \mathcal{F}_{T_1}\right)\right)\\ &= E\left(\frac{1}{B_{T_1}} P(T_1, T_2)\right)\\ &= E_1\left(\frac{dP}{dP_1}\big|_{T_1} \frac{1}{B_{T_1}} P(T_1, T_2)\right)\\ &= P(0, T_1)E_1(P(T_1, T_2))\\ &= P(0, T_1)E_1\left(\frac{1}{1+r_{12}}\right)\\ &= \frac{1}{1+r_{01}}E_1\left(\frac{1}{1+r_{12}}\right), \end{align*} where $\mathcal{F}_{T_1}$ is the information set at time $T_1$. That is, \begin{align*} \frac{1}{(1+r_{02})^2} = E_1\left(\frac{1}{1+r_{12}}\right)\frac{1}{1+r_{01}}. \end{align*}

EDIT: The last identity can also be shown by the measure change between the forward measures $P_1$ and $P_2$. Specifically, for $0 \leq t \leq T_1$, \begin{align*} \frac{dP_1}{dP_2}\big|_t = \frac{P(t, T_1)P(0, T_2)}{P(t, T_2)P(0, T_1)}. \end{align*} Consequently, \begin{align*} E_1\left(\frac{1}{1+r_{12}}\right) &= E_1\left(P(T_1, T_2)\right)\\ &= E_2\left(\frac{dP_1}{dP_2}\big|_{T_1} P(T_1, T_2)\right)\\ &= \frac{P(0, T_2)}{P(0, T_1)}\\ &= \frac{1+r_{01}}{(1+r_{02})^2}. \end{align*}

Answer 3

If the prices were not equal, there would be an immediate arbitrage opportunity as you can lock in the forward rate today. Hence the law of one price holds.