Timeline for Martingale measure result application for interest rates under T-forward measure?
Current License: CC BY-SA 3.0
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| Dec 20, 2017 at 7:28 | comment | added | Antoine Conze | 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. | |
| Dec 20, 2017 at 6:47 | comment | added | Aldo Shumway | 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? | |
| Dec 20, 2017 at 6:43 | vote | accept | Aldo Shumway | ||
| Dec 19, 2017 at 8:13 | comment | added | Antoine Conze | 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. | |
| Dec 19, 2017 at 6:35 | comment | added | Aldo Shumway | Been thinking about it and about the convexity adjustment that arises when a change of measure is applied, I have this: $p_0 = P(0, T)E^{T}\left[R(T, T, T^*) \right] = P(0, T)E^{T^*}\left[R(T, T, T^*)\frac{P(T,T)P(0,T*)}{P(T,T^*)P(0,T)} \right] = P(0, T^*)E^{T^*}\left[\frac{R(T, T, T^*)}{P(T,T*)} \right]$ $p_0 = P(0, T^*)E^{T^*}\left[\frac{R(T, T, T^*)}{P(T,T*)} \right]$ | |
| Dec 18, 2017 at 23:24 | comment | added | Aldo Shumway | 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? | |
| Dec 18, 2017 at 12:39 | history | answered | Antoine Conze | CC BY-SA 3.0 |