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I've got a question about the way the equivalent martingale measure result is used for pricing derivatives. Hull states the result as the next equality:

\begin{align*} f_o = g_0 E^{g}\big(\frac{f_T}{g_T}\mid \mathcal{F}_{t_0}\big) \end{align*}

Given that $f_T$ has some dynamics which are depedent on $g_T$ dynamics's volatility.

So what I understand is that as long as $f_T$ has the correct dynamics I can divide by $g_T$ and get the price of any derivative.

As an example, for a call with payoff $max(S_T-K,0)$ I can choose $g_0$ as the money market account with $g_0 = 1$ and $g_T = e^{rT}$ (assuming constant r). Then to price the option I would use the result like this:

\begin{align*} f_o = E^{r}\big(\frac{max(S_T-K,0)}{e^{rT}}\mid \mathcal{F}_{t_0}\big) \end{align*}

Solving this with the correct dynamics ($\mu=r$ for $S_T$) would lead us to Black and Scholes formula.

Now, in the case of interest rates I know that under a $T^*$-measure with numeraire as $P(t,T^*)$ and $T<T^*$ the forward interest rate $R(T,T,T^*)$ as seen in time $T$ is a martingale, that is:

\begin{align*} R(t_0,T,T^*) = E^{T^*}\big(R(T,T,T^*)\mid \mathcal{F}_{t_0}\big) \end{align*}

However, if I wanted to apply the same logic I used in the example before to value a derivative that pays the T-forward interest rate in time $T^*$ I would go on and do this:

\begin{align*} f_o = P(0,T^*)E^{T^*}\big(\frac{R(T,T,T^*)}{P(T,T^*)}\mid \mathcal{F}_{t_0}\big) \end{align*}

But I get the term $P(T,T^*)$ which doesn't seem right because I believe the correct valuation is:

\begin{align*} f_o = P(0,T^*)E^{T^*}\big(R(T,T,T^*)\mid \mathcal{F}_{t_0}\big)=P(0,T^*)R(t_0,T,T^*) \end{align*}

Should I be using $g_T=P(T^*,T^*)=1$ ? That doesn't seems correct to me since it's $g_T$ not $g_T^*$

I saw in here (What is the correct convexity adjustment for an Interest Rate Swap with unnatural reset lag?) and it looks like $P(T_p, T_p)$ is being used even thought the rate is observed in $T_s$

What am I missing?

Much help appreciated

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Do not confuse the fixing date $T$ and the payment date $T^*$. In your example you are valuing a floating coupon that fixes on $T$ and pays $R(T, T, T^*)$ on $T^*$, and you are using the $T^*$ zero coupon bond as numeraire, so the PV is computed as $$ p_0 = P(0, T^*)E^{T^*}\left[\frac{R(T, T, T^*)}{P(T^*,T^*)} \right] = P(0, T^*)E^{T^*}\left[R(T, T, T^*)\right]=P(0, T^*) R(0, T, T^*) $$ The floating rate $R(T, T, T^*)$ is fixed on $T$ but it is paid on $T^*$, so the denominator in the expectation is $P(T^*,T^*)$. If the floating rate was fixed AND paid on $T$ (as would be the case with an in arrears fixing) then the denominator would be $P(T,T^*)$ and that would lead to a convexity adjustment.

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  • $\begingroup$ Thanks @Antoine Conze. What confuses me it's that I'm trying to use this equation: $f_o = g_0 E^{g}\big(\frac{f_T}{g_T}\mid \mathcal{F}_{t_0}\big)$ with $g_t = P(t,T^*)$ as a numeraire which would mean that $g_T = P(T,T^*)$ and not $g_T = P(T^*,T^*)$. There's a problem however since Hull states that for this result to be true $f$ and $g$ have the next dynamics: $df = (r+ \sigma_g\sigma_f) f dt + \sigma_f f dz$ and $dg = (r+ \sigma_g^2) g dt + \sigma_g g dz$ which might not be true since $f$ is already a martingale. I'd like to see if there's a mathematical justification for this calculation? $\endgroup$
    – Aldo Shumway
    Commented Dec 18, 2017 at 23:24
  • $\begingroup$ Been thinking about it and about the convexity adjustment that arises when a change of measure is applied, I have this: $p_0 = P(0, T)E^{T}\left[R(T, T, T^*) \right] = P(0, T)E^{T^*}\left[R(T, T, T^*)\frac{P(T,T)P(0,T*)}{P(T,T^*)P(0,T)} \right] = P(0, T^*)E^{T^*}\left[\frac{R(T, T, T^*)}{P(T,T*)} \right]$ $p_0 = P(0, T^*)E^{T^*}\left[\frac{R(T, T, T^*)}{P(T,T*)} \right]$ $\endgroup$
    – Aldo Shumway
    Commented Dec 19, 2017 at 6:35
  • $\begingroup$ if you use $g_T = P(T, T^*)$ that means you are pricing a cash flow that pays on $T$, not on $T^*$, because $P(T, T^*)$ is in time $T$ money (it is the time $T$ value of the $T^*$ maturity zero coupon bond). Hence your confusion. More generally if the cash flow pays on $T_p$ and you are using the $T^*$-forward measure then the denominator is $P(T_p, T^*)$. It goes in finance as in physics, always check that your units are consistent. $\endgroup$
    – Antoine Conze
    Commented Dec 19, 2017 at 8:13
  • $\begingroup$ Thanks, I think now I get what you are saying, so I can choose the same numéraire $g_t = P(t, T^*)$ but change the $t$ according to the time of payment. Whereas if I choose $h_t = P(t, T)$ that would be a different numéraire that could lead me to a convexity adjustment. I think the problem is that the equation Hull presents doesn't allow for a distintion with the time of payment and the time of observation. I like it because he gives a simple proof of how the process $f/g$ becomes a martingale, Do you know any reference where I could find another proof of the martingale measure result? $\endgroup$
    – Aldo Shumway
    Commented Dec 20, 2017 at 6:47
  • $\begingroup$ The original papers on the martingale measure approach are J.M. Harrison and D. Kreps. Martingales and arbitrage in multiperiod securities markets. Journal of Economic theory, 20(3) :381–408, 1979. and J.M. Harrison and S.R. Pliska. Martingales and stochastic integrals in the theory of conti- nuous trading. Stochastic processes and their applications, 11(3) :215–260, 1981. $\endgroup$
    – Antoine Conze
    Commented Dec 20, 2017 at 7:28

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