6
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Dec 18 '17 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 Dec 19 '17 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 Dec 19 '17 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 Dec 20 '17 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 Dec 20 '17 at 7:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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