# Equality under T-forward measure for convexity adjustment

I've been working with the convexity adjustment for interest rates that arises when changing from one measure $Q_{T_p}$ with a numéraire $N_p=P(t,T_p)$ to a measure $Q_{T_e}$ with a numéraire $N_e=P(t,T_e)$ where $T_p$ is the time of payment and $T_e$ is the time where the interest of the forward rate ends.

So I have the forward rate:

\begin{align*} L(t, T_s, T_e) = \frac{1}{\Delta_s^e}\bigg(\frac{P(t, T_s)}{P(t, T_e)}-1 \bigg), \end{align*}

And the expecation of the payment with the change of measure:

\begin{align*} &\ P(t_0, T_p)E^{T_p}\big(L(T_s, T_s, T_e) \mid \mathcal{F}_{t_0}\big) \\ =&\ P(t_0, T_p)E^{T_e}\Big(\frac{\eta_{T_p}}{\eta_{t_0}}L(T_s, T_s, T_e) \mid \mathcal{F}_{t_0}\Big)\\ =&\ P(t_0, T_p)E^{T_e}\Big(\frac{P(t_0, T_e)}{P(t_0, T_p)P(T_p, T_e)} L(T_s, T_s, T_e)\mid \mathcal{F}_{t_0}\Big)\\ =&\ P(t_0, T_e)E^{T_e}\Big(\frac{1}{P(T_p, T_e)} L(T_s, T_s, T_e)\mid \mathcal{F}_{t_0}\Big)\\ =&\ P(t_0, T_e)E^{T_e}\Big(\big(1+ \Delta_p^e L(T_p, T_p, T_e) \big) L(T_s, T_s, T_e)\mid \mathcal{F}_{t_0}\Big) \end{align*}

Now I believe, this last equation is the tricky one since I have two forward rates observed in times $T_p$ and $T_s$ I'd like to confirm my thinking that since we are under the $Q_{T_e}$ measure and both $L(T_p, T_p, T_e)$ and $L(T_s, T_p, T_e)$ are martingales under it, I can use this equality:

\begin{align*} P(t_0, T_e)E^{T_e}\Big(\big(1+ \Delta_p^e L(T_p, T_p, T_e) \big) L(T_s, T_s, T_e)\mid \mathcal{F}_{t_0}\Big)= P(t_0, T_e)E^{T_e}\Big(\big(1+ \Delta_p^e L(T_s, T_p, T_e) \big) L(T_s, T_s, T_e)\mid \mathcal{F}_{t_0}\Big)\tag{1} \end{align*}

Does this equality holds because that reason? Now from there I've seen two possible solutions:

1. Assume a model (e.g. log-normal) for both $L(t, T_p, T_e)$ and $L(t, T_p, T_e)$ since now we'll be working with both values observed in $T_s$ when I use Ito's lemma to compute the product and then integrate I'll do it from $t_0$ to $T_s$ and with that solution can solve the expectation.

2. Use the linear model proposed here in page 19,Section 4.2. where essentially the calculations are made with respect to $(1+ \Delta_s^e L(T_s, T_s, T_e)$ but with a specific value to the $\Delta_s^e$ being multiplied in order to account for the payment being in $T_p$ instead of in $T_e$.

I'm working on this to price an option, specifically a digital option (which I think should be almost the same). Have anyone here used any of this results of would have a preference over one of these options?

Much help appreciated

So in your particular case, to price a digital option that pays $\text{Indicator}(L(T_s, T_s, T_e) > K)$ on $T_p$, you first apply replication to price options with payoff $(L(T_s, T_s, T_e) - K)^+$ that pay on $T_p$, and you then differentiate (numerically) with respect to $K$.
• Thanks @Antoine Conze, I've been reading about what you commented and I've got a couple of questions, so the idea for pricing this delayed payoff is to make a change of measure which introduces this factor $\frac{1}{P(T_p, T_e)}$ in the expectation and the one chooses a way of approximating this factor in terms of $L(T_s, T_s, T_e)$, might be the one I posted in eq (1) with $L(T_p, T_p, T_e) =L(T_s, T_s, T_e)$ or a better one. Then one calculates the expectation applying replication because I guess that gives us a more realistic model than assuming log-normality Dec 25, 2017 at 18:59
• (I'm assuming that's the main reason because I haven't found more detail about it yet). And finally for a digital option I'd estimate $-\frac{\partial call}{\partial K}$ to get the needed payoff. Are these the "steps" for the valuation? Because it seems right to me, only the last part bugs me, why the need of differentiating? can't the expecation be calculated directly with replication and still be consistent with the smile by choosing the correct implied volatility according to the strike? Also, thanks for the book reference and happy holidays Dec 25, 2017 at 19:27
• Yes these are the right steps. As for the digital option $E^{T_p}[Indicator(L(T_s,T_s,T_e)>K)]$ you can indeed obtain it by replication but the usual approach is to use differentiation of $E^{T_p}[(L(T_s,T_s,T_e)-K)^+]$ wrt $K$, because the latter is the standard approach for natural payment date options (compute the vanilla option from the vol at the corresponding strike, then differentiate the vanilla option price wrt strike). Dec 26, 2017 at 8:31
• Assume payoff $X_{T_s}$ is measurable wrt filtration at time $T_s$. From conditional expectation rule, $E^{T_e}[X_{T_s}\frac{1}{P(T_p,T_e)}] = E^{T_e}[X_{T_s} E^{T_e}[\frac{1}{P(T_p,T_e)}|\mathcal{F}_{T_s}]]$. Now because $Q^{T_e}$ is risk neutral wrt maturity $T_e$ zero coupon bond, $E^{T_e}[\frac{1}{P(T_p,T_e)}|\mathcal{F}_{T_s}] = \frac{P(T_s, T_p)}{P(T_s, T_e)}$. Dec 29, 2017 at 7:46