Frequently I come across the statement that the "correct measure" for a product is this-or-that measure. For example,

  1. Eurodollar Futures or Stock returns - Risk neutral measure
  2. Libor forward rate - T-forward measure
  3. Libor in arrears - T-1-forward measure etc.
  4. Swaption - annuity measure

The explanation for this is that the payoff is a martingale under this measure. But I do not understand the logic here - are we making an assumption about the martingale property, based on some reasonable justification? How does one go about finding the right measure for a product?

  • $\begingroup$ is it still unclear? Is there anything you do not understand? $\endgroup$ Commented Feb 27, 2020 at 18:05
  • $\begingroup$ @Bravo: if you like Daneel's (or mine) answer below, could you pls click on the "tick mark" next to one of the answers so that this question can be marked as "answered"? $\endgroup$ Commented Jun 9, 2020 at 16:12

3 Answers 3


Recall that any traded asset divided by a numéraire is a martingale under the measure associated to that numéraire. For the 3 interest rates you mention, the natural measure (namely the one that makes those processes martingales) is deduced from the structure of the rate.

Always keep in mind that the value at $t_0$ of a cash flow $C$ paid at $T$ is equal to the conditional expectation of the discounted cash flow under the risk-neutral measure $\mathcal{Q}$: $$C(t_0)=E_{t_0}^\mathcal{Q}\left(\frac{B_{t_0}}{B_T}C(T)\right)$$ where $B_t$ is the money market account: $B_t=e^{rt}$, namely the numéraire under the risk-neutral measure. You thus notice that: $$D(t_0,T)=\frac{B_{t_0}}{B_T}$$ where $D(t_0,T)$ is the discount factor from $t_0$ to $T$. The theory of numéraire change developed by Geman, El Karoui and Rochet in [1] establishes the following equivalency between measures: $$E_{t_0}^\mathcal{Q}\left(\frac{B_{t_0}}{B_T}C(T)\right) =E_{t_0}^\mathcal{N}\left(\xi(t_0,T)\frac{B_{t_0}}{B_T}C(T)\right) =E_{t_0}^\mathcal{N}\left(\frac{N_{t_0}}{N_T}C(T)\right)$$ where $N_t$ is another asset which might be used as a numéraire, $\mathcal{N}$ its associated measure, and $\xi(t_0,T)$ the associated Radon-Nikodym derivative to change from one measure to the other: $$\xi(t_0,T)=\frac{B_TN_{t_0}}{B_{t_0}N_T}$$

Forward LIBOR rate: you probably know that the forward LIBOR rate is equal to: $$L(t,T,T+\delta)=\frac{1}{\delta}\left(\frac{P(t,T)}{P(t,T+\delta)}-1\right)$$ where $P(t,T)$ is a zeron-coupon bond with maturity $T$. Now, such a product is a traded asset with a strictly positive price and no dividends, therefore it can be used as a numéraire. Hence under the $T+\delta$ measure the LIBOR rate is a martingale: $$\begin{align} E_{t_0}^{T+\delta}\left(L(t,T,T+\delta)\right) &=E_{t_0}^{T+\delta}\left(\frac{1}{\delta}\left(\frac{P(t,T)}{P(t,T+\delta)}-1\right)\right) \\ &=\frac{1}{\delta}\left(E_{t_0}^{T+\delta}\left(\frac{P(t,T)}{P(t,T+\delta)}\right)-1\right) \\ &=\frac{1}{\delta}\left(\frac{P(t_0,T)}{P(t_0,T+\delta)}-1\right) \\[7pt] &=L(t_0,T,T+\delta) \end{align}$$

LIBOR-in-arrears: in this case you are paid at $T$ the LIBOR for the period $[T,T+\delta]$. There is no measure under which the LIBOR-in-arrears is a pure martingale. The value of this flow is: $$\begin{align} E_{t_0}^\mathcal{Q}\left(\frac{B_{t_0}}{B_T}L(T,T,T+\delta)\right) &=P(t_0,T)E_{t_0}^T\left(L(T,T,T+\delta)\right) \\ &=P(t_0,T+\delta)E_{t_0}^{T+\delta}\left(\frac{L(T,T,T+\delta)}{P(T,T+\delta)}\right) \\[4pt] &=P(t_0,T+\delta)\left(L(t_0,T,T+\delta)+\delta E_{t_0}^{T+\delta}\left( L(T,T,T+\delta)^2\right)\right) \end{align}$$ where the term $\delta E_{t_0}^{T+\delta}(L(T,T,T+\delta)^2)$ is a convexity adjustment.

Swap rate: the value of the swap rate $S(t_0)$ at time $t_0$ is derived by setting equal the values of the two legs of a swap starting at $t_0$, namely: $$\sum_{i=1}^n\delta_i^SS(t_0)P(t_0,t_i)=\sum_{j=1}^m\delta_j^LL(t_0,t_{i-1},t_i)P(t_0,t_i)$$ Rearranging: $$S(t_0)=\frac{\sum_{j=1}^m\delta_j^LL(t_0,t_{i-1},t_i)P(t_0,t_i)}{A^S(t_0,t_n)}$$ where the swap annuity $A(t_0,t_n)$ is defined as: $$A^S(t_0,t_n)=\sum_{i=1}^n\delta_i^SP(t_0,t_i)$$ The annuity is a portfolio of zero-coupon bonds (traded assets), thus it can be used as a numéraire. You therefore see that under the measure $\mathcal{A}$ associated to the annuity, the swap rate will be a martingale by a similar argument to the forward LIBOR rate: $$\begin{align} E_{t_0}^\mathcal{A}\left(S(t)\right) &=E_{t_0}^\mathcal{A}\left(\frac{\sum_{j=1}^m\delta_j^LL(t,t_{i-1},t_i)P(t,t_i)}{A^S(t,t_n)}\right) \\ &=\frac{\sum_{j=1}^m\delta_j^LL(t_0,t_{i-1},t_i)P(t_0,t_i)}{A^S(t_0,t_n)} \\[7pt] &=S(t_0) \end{align}$$

The forward LIBOR and swap rate cases clearly show that the proper martingale measure depends on the structure of the product being considered.

Note also that products like swaptions are quoted on a Bachelier/Black implied-volatility basis, that is the swaption is quoted with the implied volatility that corresponds to its market price. This implied volatility is obtained by inverting the Bachelier/Black formulas through numerical methods. Now, these formulas assume the underlying market factor (i.e. the swap rate) is a martingale under the pricing measure, thus the term “natural measure”: it is the measure implied by the market’s quotation convention.


[1] Geman, H., El Karoui, N., Rochet, J.C. (1995). "Changes of Numéraire, Changes of Probability Measures and Pricing of Options", on Journal of Applied Probability, Vol. 32, pg 443-458.

  • 1
    $\begingroup$ The def of martingale is: $E(M_t\mid F_{t_0})=M_{t_0}$. How can we say: $E_{t_0}^{T+\delta}\frac{1}{P(t,T+\delta)}=\frac{1}{P(t_0,T+\delta)}$, as the function is in the denominator (won't Jensen's inequality apply?) $\endgroup$
    – Bravo
    Commented Feb 25, 2020 at 11:21
  • $\begingroup$ The theory of change of numéraire demonstrates that the asset divided by the numéraire is a martingale. In this case, the numéraire is the zero-coupon bond with maturity $T+\delta$, i.e. $P(\cdot,T+\delta)$. $\endgroup$ Commented Feb 25, 2020 at 11:26
  • $\begingroup$ Note that the equation in your comment is not correct: an asset with a payoff of $1$ at some time $t’$ corresponds to a zero-coupon bond with maturity $t’$, thus its price at $t\leq t’$ is given by $P(t,t’)$ with $P(t’,t’)=1$. $\endgroup$ Commented Feb 25, 2020 at 11:29
  • $\begingroup$ So if your asset is $X(t)$ and your numéraire is $N(t)$, it is the process $X(t)/N(t)$ which is a martingale under the measure $\mathcal{N}$ induced by the numéraire $N(t)$. $\endgroup$ Commented Feb 25, 2020 at 11:31
  • $\begingroup$ I don’t understand why you mention Jensen’s inequality. $\endgroup$ Commented Feb 25, 2020 at 11:33

I would like to add to @DaneelOlivaw answer.

Your question: "How does one go about finding the right measure for a product?"

Answer: One should choose any measure that will make it easy and convenient to compute the pricing at hand.

We are free to use whichever measure we would like. For example, it is possible to derive the process for the Forward Libor $L(t,T_1,T_2)$ under a different measure than the one associated with $P(t,T_2)$ as Numeraire. However, such process would be a lot more complicated. So if we were to price options on Forward Libor under a different measure, we would make things unnecessarily more complex for ourselves.

Specific example:

$$1 + \delta L(t,T_1,T_2) = \frac{P(t,T_1)}{P(t,T_2)}$$


$$L(t,T_1,T_2) = \frac{1}{\delta} \left( \frac{P(t,T_1)-P(t,T_2}{P(t,T_2)}\right)$$


$$L(t,T_1,T_2)P(t,T_2) = \frac{1}{\delta} \left( P(t,T_1)-P(t,T_2) \right)$$

We know the right-hand side is a linear combination of traded assets (i.e. zero coupon bonds with different maturities) so we know that these have to be a Martingale under a numeraire of our choice. Chose $P(t,T_1)$ as numeraire:

$$\mathbb{E}^{P_{T(1)}} \left[ \frac{1}{\delta} \frac{P(t,T_1)-P(t,T_2)}{P(t,T_1)} \right] = martingale = \mathbb{E}^{P_{T(1)}} \left[ \frac{L(t,T_1,T_2)P(t,T_2)}{P(t,T_1)} \right] $$

Notice that on the RHS, we have the Libor process $L(t,T_1,T_2)$ multiplied by the bond $P(t,T_2)$ and divided by the bond $P(t,T_1)$ and this whole expression has to be a martingale for no-arbitrage pricing: so the above is not very helpful in the sense that we now have to worry about coming up with mathematical processes for $L(t,T_1,T_2)$, $P(t,T_2)$ and $P(t,T_1)$ such that their fraction is a martingale.

But, what if, instead of using $P(t,T_1)$ as Numeraire, we decide to use $P(t,T_2)$ as Numeraire?

$$\mathbb{E}^{P_{T(2)}} \left[ \frac{1}{\delta} \frac{P(t,T_1)-P(t,T_2)}{P(t,T_2)} \right] = martingale = \\ = \mathbb{E}^{P_{T(2)}} \left[ \frac{L(t,T_1,T_2)P(t,T_2)}{P(t,T_2)} \right] = \mathbb{E}^{P_{T(2)}} \left[ L(t,T_1,T_2)\right]$$

We can now directly deduce the process for $L(t,T_1,T_2)$ (using $P(t,T_2)$ as Numeraire) as:

$$ L(t,T_1,T_2)=L(t_0,T_1,T_2)exp\left( -0.5 \sigma^2t + \sigma W(t) \right) $$

Because we know that under the $P(t,T_2)$ Numeraire, $L(t,T_1,T_2)$ alone must be a martingale.

We could have chosen $P(t,T_1)$ as Numeraire, but we'd have made things a lot more difficult for ourselves (just to stress the point again: because we'd have to think out the process for $L(t,T_1,T_2)$ such that $\frac{L(t,T_1,T_2)P(t,T_2)}{P(t,T_1)}$ is a martingale, rather than just $L(t,T_1,T_2)$ being a martingale).

Conclusion: the change of measure technique is all about convenience and computability. It is a mathematical technique that allows one to simplify the pricing task at hand.

  • $\begingroup$ If anyone has any other good examples of why a certain change of measure is particularly helpful, in addition to the examples already posted, please do share: I feel this is a relevant topic. For example, any specific examples of when the measure associated with the Stock-price Numeraire is particularly beneficial? $\endgroup$ Commented Jun 8, 2020 at 18:00
  • 1
    $\begingroup$ Strictly speaking unrelated to derivative pricing, but here is a nice example: Probability in different measures. $\endgroup$ Commented Jun 11, 2020 at 8:25

Via a combination of the Cameron-Martin-Girsanov Theorom and the Martingale Representation Theorom you can find the equivalent martingale measure.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.