I am trying to prove the need of a convexity adjustment to a forward rate by calculating the next expectation:

\begin{align*} P(t_0, T_s)E^{T_s}\big(L(T_s, T_s, T_e) \mid \mathcal{F}_{t_0}\big). \end{align*}

Where $E^{T_s}$ denotes the expectation under a T-measure with $P(t,T_s)$ as numéraire and $t_0< T_s < T_e $ and $L(T_s, T_s, T_e)$ is the libor rate observed in $T_s$ for the period between $T_s$ and $T_e$

To do it I would like to apply a change of measure so that I can calculate the expectation under a T*-measure with $P(t,T_e)$ as numéraire.

I know to do this change of measure I need to know the Radon-Nikodym derivative, so I need something like this:

\begin{align*} P(t_0, T_s)E^{T_s}\big(L(T_s, T_s, T_e) \mid \mathcal{F}_{t_0}\big)=P(t_0, T_s)E^{T_e}\big(\frac{dQ^{T_s}}{dQ^{T_e}}L(T_s, T_s, T_e) \mid \mathcal{F}_{t_0}\big) \end{align*} How do I know what value of $\frac{dQ^{T_s}}{dQ^{T_e}}$ changes from $Q^{T_s}$ to $Q^{T_e}$?

From what I've seen so far, the Radon-Nikodym derivative is easy to get when you have the distribution under which you are trying to calculate the expectation. For example if $X \sim N(0,1)$ with density function $f(x)$ I can calculate $E[X]$ the usual integral way, or I can introduce a measure $G$ where $g(x)$ can be the density function of say $X \sim N(0,100)$ and it would be the same if I calculate $E_g[X\frac{f(x)}{g(x)}]$ so here my Radon-Nikodym derivative is the division of two density functions. I've seen different publications in where this is used to change from one measure to another, but still I don't seem to understand how you know what value to use for each case, specially in the case I'm asking now since I'm not sure of the density functions I should be using.

The only thing that cross through my mind is that $L(T_s, T_s, T_e)$ is a martingale under $Q^{T_e}$. So perhaps I should assign it this dynamics $dL(t, T_s, T_e) = \sigma_s L(t, T_s, T_e) d W_t^s$ from there I can get a density function which would be like the $g(x)$ in my example. Then if I can find how $L(T_s, T_s, T_e)$ dynamics are under $Q^{T_s}$ maybe I could get the $f(x)$ and the division would be my Radon-Nikodym?

Much help appreciated


3 Answers 3


By definition $Q^{T_s}$ is risk neutral for the numeraire $P(t,T_s)$, and $Q^{T_e}$ is risk neutral for the numeraire $P(t,T_e)$, hence $$ \left(\frac{dQ^{T_s}}{dQ^{T_e}}\right)_t = \frac{P(t,T_s)}{P(t,T_e)} \frac{P(t_0,T_e)}{P(t_0,T_s)} $$ In the specific case that you are looking at you are computing the forward in-arrears fixing Libor (in arrears because fixed and paid on $T_s$) so what you need is $$ \left(\frac{dQ^{T_s}}{dQ^{T_e}}\right)_{T_s} = \frac{P(T_s,T_s)}{P(T_s,T_e)} \frac{P(t_0,T_e)}{P(t_0,T_s)} = \frac{1}{P(T_s,T_e)} \frac{P(t_0,T_e)}{P(t_0,T_s)} $$ In a single curve settings you have by definition of the Libor rate $$ P(T_s,T_e) = \frac{1}{1+L(T_s, T_s, T_e) \text{yearfrac}(T_s,T_e)} $$ hence $$ \left(\frac{dQ^{T_s}}{dQ^{T_e}}\right)_{T_s} =\left(1+L(T_s, T_s, T_e) \text{yearfrac}(T_s,T_e)\right) \frac{P(t_0,T_e)}{P(t_0,T_s)} $$ and $$ E^{T_s}\left[L(T_s, T_s, T_e) \right] = \frac{P(t_0,T_e)}{P(t_0,T_s)} E^{T_e}\left[L(T_s, T_s, T_e) \left(1+L(T_s, T_s, T_e) \text{yearfrac}(T_s,T_e)\right)\right] \\ = E^{T_e}\left[L(T_s, T_s, T_e) \right] + cvx $$ with $$ cvx = \frac{P(t_0,T_e)}{P(t_0,T_s)} E^{T_e}\left[L(T_s, T_s, T_e) \left(1+L(T_s, T_s, T_e) \text{yearfrac}(T_s,T_e)- \frac{P(t_0,T_s)}{P(t_0,T_e)} \right)\right] $$ This is the theoretical convexity adjustment.

To compute the adjustment you need a model for $L(T_s, T_s, T_e)$. For instance if you assume that $L(T_s, T_s, T_e)$ is log normal or displaced log normal with constant volatility you easily obtain a closed form solution.

Or if you assume that prices of caplets/floorlets on $L(T_s, T_s, T_e)$ with natural payment date $T_e$ are available for all strikes you can compute the adjustment using replication and the Carr-Madan formula. The latter is the standard procedure for in-arrears swaps / caps / floors.

In a dual curve settings you can easily adapt the above formulas by assuming for instance that the Libor-OIS basis is deterministic.

Also in real life for most markets (notable exception is GBP) a Libor that covers the period $T_s$ to $T_e$ fixes on $T_s - 2$ business days, but the approach above still applies.

  • $\begingroup$ hey, @Antoine Conze thanks for the info. I'm still lost with the Radon-Nikodym derivative part, could you give me more insight on it? how do you calculate it? My problem is that I'm looking for probabilities in the RHS of the equation but all I'm understanding is that I have discount factors $\left(\frac{dQ^{T_s}}{dQ^{T_e}}\right)_t = \frac{P(t,T_s)}{P(t,T_e)} \frac{P(t_0,T_e)}{P(t_0,T_s)}$ $\endgroup$ Commented Dec 7, 2017 at 6:50
  • $\begingroup$ RN derivative is just a random variable: $dQ^{T_s}/dQ^{T_e} = Z$ means $E^{T_s}[X] = E^{T_e}[Z X]$ for any $X$. You don't necessarily have to introduce the density of $Z$ to perform calculations. $\endgroup$ Commented Dec 7, 2017 at 8:11
  • $\begingroup$ thanks again. @Antoine Conze I get that RN derivative helps us move from one probability measure to another. What I don't understand is how you know that in this case $\left(\frac{dQ^{T_s}}{dQ^{T_e}}\right)_t$ equals $\frac{P(t,T_s)}{P(t,T_e)} \frac{P(t_0,T_e)}{P(t_0,T_s)}$ ? that's the part I can't figure out. Is it that for some reason $dQ^{T_s}= \frac{P(t,T_s)}{P(t,T_e)}$ and $dQ^{T_e}= \frac{P(t_0,T_s)}{P(t_0,T_e)}$ ? $\endgroup$ Commented Dec 7, 2017 at 20:02
  • $\begingroup$ I kept searching and found on here that radon-Nikodym derivative is given by the ratio of the numeraires. Is that what you are using to calculate it? $\endgroup$ Commented Dec 7, 2017 at 22:04
  • 1
    $\begingroup$ yes you are correct. An additional check is that one should recover cvx = 0 when rates are deterministic. Thank you for pointing it out. $\endgroup$ Commented Mar 21, 2021 at 15:59

I think that your question can be solved easier. You may ask me why. Here is my answer:

First of all the LIBOR forward rate $L(t, t, T)$ is $\mathbb{Q}^{T}$-martingale, where $\mathbb{Q}^{T}$ is a $T-$forward measure defined with the following Ranon-Nikidym derivative structure:

\begin{equation} \displaystyle\frac{d\mathbb{Q}^T}{d\mathbb{P}} = \frac{e^{-\int_{0}^{T}\, r_u du}}{P(0, T)} \end{equation}

Therefore, using the standard definition for the spot forward LIBOR rate we have that

\begin{equation} P(0, T)\mathbb{E}^{\mathbb{Q}^T}\Big(L(t,t, T)\Big) = P(0, t)\times L(0, t, T) = P(0, t)\times\frac{1}{\Delta}\Big(\frac{P(0, t)}{P(0, T)}-1\Big), \end{equation} where $\Delta = T-t$, and $P(0, t)$ and $P(0, T)$ are zero-coupon bond prices with different maturity times.


I am actually getting a slightly different convexity adjustment to Antoine's:

For clarity of notation, I use: $T_s=T_1$, $T_e=T_2$ and $yearfrac(T_1,T_2)=\tau$.

We then have (by definition of Radon-Nikodym derivative):

$$\mathbb{E}^{Q_{T_1}}_{t_0}\left[L(T_1, T_1, T_2)\right]=\mathbb{E}^{Q_{T_2}}_{t_0}\left[\frac{\partial Q_{T_1}}{\partial Q_{T_2}}L(T_1, T_1, T_2)\right]$$

The Radon-Nikodym derivative is then computed as:

$$\frac{\partial Q_{T_1}}{\partial Q_{T_2}}(T_1)|t_0=\frac{P(t_0,T_2)}{P(t_0,T_1)}\frac{P(T_1,T_1)}{P(T_1,T_2)}=\frac{P(t_0,T_2)}{P(t_0,T_1)}(1+\tau L(T_1,T_1,T_2))$$

(so far our results agree)

And substituting the above into the expectation, we get:

$$\mathbb{E}^{Q_{T_2}}_{t_0}\left[\frac{\partial Q_{T_1}}{\partial Q_{T_2}}L(T_1, T_1, T_2)\right]=\mathbb{E}^{Q_{T_2}}_{t_0}\left[\frac{P(t_0,T_2)}{P(t_0,T_1)}(1+\tau L(T_1,T_1,T_2))L(T_1, T_1, T_2)\right]=\\=\mathbb{E}^{Q_{T_2}}_{t_0}\left[\frac{P(t_0,T_2)}{P(t_0,T_1)}L(T_1,T_1,T_2)+\frac{P(t_0,T_2)}{P(t_0,T_1)}\tau L(T_1,T_1,T_2)^2\right]=\\=$$

Now adding $0=+\mathbb{E}^{Q_{T_2}}_{t_0}\left[L(T_1,T_1,T_2)\right]-\mathbb{E}^{Q_{T_2}}_{t_0}\left[L(T_1,T_1,T_2)\right]$, we get:

$$=\\=\mathbb{E}^{Q_{T_2}}_{t_0}\left[\frac{P(t_0,T_2)}{P(t_0,T_1)}L(T_1,T_1,T_2)+\frac{P(t_0,T_2)}{P(t_0,T_1)}\tau L(T_1,T_1,T_2)^2\right]+\mathbb{E}^{Q_{T_2}}_{t_0}\left[L(T_1,T_1,T_2)\right]-\mathbb{E}^{Q_{T_2}}_{t_0}\left[L(T_1,T_1,T_2)\right]=\\=\mathbb{E}^{Q_{T_2}}_{t_0}\left[\frac{P(t_0,T_2)}{P(t_0,T_1)}L(T_1,T_1,T_2)+\frac{P(t_0,T_2)}{P(t_0,T_1)}\tau L(T_1,T_1,T_2)^2-L(T_1,T_1,T_2)\right]+\mathbb{E}^{Q_{T_2}}_{t_0}\left[L(T_1,T_1,T_2)\right]=\\=\mathbb{E}^{Q_{T_2}}_{t_0}\left[\frac{P(t_0,T_2)}{P(t_0,T_1)}L(T_1,T_1,T_2)\left(1+\tau L(T_1,T_1,T_2)-\frac{P(t_0,T_1)}{P(t_0,T_2)} \right)\right]+\mathbb{E}^{Q_{T_2}}_{t_0}\left[L(T_1,T_1,T_2)\right]=\\=\mathbb{E}^{Q_{T_2}}_{t_0}\left[L(T_1,T_1,T_2)\right]+cvx$$

From the above, it follows that:

$$cvx=\mathbb{E}^{Q_{T_2}}_{t_0}\left[\frac{P(t_0,T_\color{red}2)}{P(t_0,T_\color{red}1)}L(T_1,T_1,T_2)\left(1+\tau L(T_1,T_1,T_2)-\frac{P(t_0,T_\color{red}1)}{P(t_0,T_\color{red}2)} \right)\right]$$

As opposed to:

$$cvx=\mathbb{E}^{Q_{T_2}}_{t_0}\left[L(T_1,T_1,T_2)\left(1+\tau L(T_1,T_1,T_2)-\frac{P(t_0,T_\color{red}2)}{P(t_0,T_\color{red}1)} \right)\right]$$


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.