Change of measure between T-forward and T*-forward contract?

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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:

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