Gaussian Path Integrals for Zero-Coupon Bond Call Options
Summary
The document works through a proposed Gaussian path-integral calculation arising in a derivation of a call option on a zero-coupon bond. It applies the generating functional for Brownian motion to a source containing both a real term and an imaginary term proportional to a Fourier variable. Expanding the square yields a quadratic term in that variable and a cross term involving the two volatility integrals.
The author uses identities relating the volatility and drift functions to simplify the expression, but gets stuck on the factor of the imaginary unit in the cross term. As presented, the identities do not directly remove that factor: the third identity contains a cross term without the Fourier variable, while the expanded expression contains it multiplied by the variable. The derivation therefore exposes a possible missing condition or mismatch in the stated identities. It is a calculation question rather than a completed pricing formula, and assumes the functions are sufficiently well behaved.
Key ideas
- The Brownian generating functional turns a linear source into an exponential quadratic in that source.
- A complex source produces a cross term proportional to the Fourier variable and the imaginary unit.
- The stated drift-volatility identities do not by themselves eliminate that Fourier-dependent cross term.
- The calculation is an intermediate step in a path-integral approach to pricing a bond call option.
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# Path integral approach to price call option on zero coupon bonds
# Path integral approach to price call option on zero coupon bonds
I am given the following identities: $$ Z[J,t_1,t_2]=\int D W e^{\int_{t_1}^{t_2}dtJ(t)W(t)}e^{S}=e^{\frac{1}{2}\int_{t_1}^{t_2}dtJ(t)^2} $$ $$ \int_t^Tdx\alpha(t,x)=\frac{1}{2}\left[\int_t^Tdx\sigma(t,x)\right]^2 $$ $$ \int_{t_0}^{t_\ast}dt\left[\int_{t_\ast}^Tdx\alpha(t,x)-\int_t^{t_\ast}dx\sigma(t,x)\int_{t_\ast}^Tdy\sigma(t,y)\right]=\frac{1}{2}\int_{t_0}^{t_\ast}dt\left[\int_{t_\ast}^Tdx\sigma(t,x)\right]^2 $$ and the path integral $$ \int D W e^{-\int_{t_0}^{t_\ast}dt\int_t^{t_\ast}dx\sigma(t,x)W(t)+ip\int_{t_0}^{t_\ast}dt\int_{t_\ast}^Tdx\sigma(t,x)W(t)}e^S. $$ I aim to simplify the path integral above by using the identities. The expression I am supposed to arrive at is $$ \exp\left[-\frac{p^2}{2}\int_{t_0}^{t_\ast}dt\left[\int_{t_\ast}^Tdx\sigma(t,x)\right]^2+\int_{t_0}^{t_\ast}dt\int_{t}^{t_\ast}dx\alpha(t,x)\right]. $$ Now, the first exponential in the path integral's integrated can be written in the form of the first identity: $$ \exp\left[\int_{t_0}^{t_\ast}dtW(t)\left(-\int_t^{t_\ast}dx\sigma+ip\int_{t_\ast}^Tdx\sigma\right)\right], $$ where we left the dependence of $\sigma$ on $t,x$ implicit as simplification. Hence, integrating over $W$, using the first identity $$ \exp\left[\frac{1}{2}\int_{t_0}^{t_\ast}dt\left(-\int_t^{t_\ast}dx\sigma+ip\int_{t_\ast}^Tdx\sigma\right)^2\right]. $$ Working out the square in the exponential gives $$ \left(\int_t^{t_\ast}dx\sigma\right)^2-p^2\left(\int_{t_\ast}^Tdx\sigma\right)^2-2ip\int_t^{t_\ast}dx\sigma(t,x)\int_{t_\ast}^Tdy\sigma(t,y). $$ Using the second identity, this is equal to $$ 2\int_t^{t_\ast}dx\alpha(t,x)-2p^2\int_{t_\ast}^Tdx\alpha(t,x)-2ip\int_t^{t_\ast}dx\sigma(t,x)\int_{t_\ast}^Tdy\sigma(t,y) $$
Plugging this back into the exponential yields $$ e^{\int_{t_0}^{t_\ast}dt\int_{t}^{t_\ast}dx\alpha(t,x)}\exp\left[-\int_{t_0}^{t_\ast}dt\left(p^2\int_{t_\ast}^Tdx\alpha(t,x)+ip\int_t^{t_\ast}dx\sigma(t,x)\int_{t_\ast}^Tdy\sigma(t,y)\right)\right]. $$ This is the furthest I have gotten. Although the second exponential above very much resembles the third identity, I can not figure out how to get rid of the factor "$ip$" in the second term.
PS: For those interested, this is part of a derivation of the price of a (financial) call option on a zero coupon bond using a path integral approach proposed in the article "Quantum Field Theory of Treasury Bonds" by Baaquie (2001).
PPS: This question is purely calculation related. One can assume that all the relevant functions behave properly for all the mathematics to make sense.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.