Interpreting a Jump-Diffusion SDE for Forward Prices
Summary
The document explains how to interpret a forward-price stochastic differential equation containing both Brownian motion and Poisson jumps. It treats the counting-process increment as zero between event times and one at a jump, which lets the exponential jump expression be rewritten as a jump-size coefficient multiplied by the counting-process increment. The resulting price process has a continuous diffusion component, a drift component, and discrete price jumps.
To clarify the jump behavior, the answer applies Itô’s formula for semimartingales with jumps to the logarithm of the forward price. At a jump, the price changes by a multiplicative factor, so the log price changes by the corresponding log jump size. This gives a log-price equation that can be integrated term by term. The explanation assumes a standard Poisson process and states that it is independent of the Brownian motion; it does not develop the drift specification or discuss calibration and model fit.
Key ideas
- A Poisson counting-process increment is zero between jumps and records an event at a jump time.
- The exponential jump term can be expressed as a jump-size coefficient multiplied by the counting-process increment.
- At a jump, the forward price changes multiplicatively while its logarithm changes additively.
- Itô’s formula for processes with jumps separates the log-price dynamics into diffusion, drift, and jump components.
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# How is this SDE interpreted?
# How is this SDE interpreted?
I saw this model
$$\frac{dF(t,T)}{F(t,T)} = \sigma(t,T) dW_t + (\exp(e^{-a(T-t)}dJ_t)-1) + \mu_J(t,T)dt$$
to model the forward curve. Rewriting
$$dF(t,T) = \sigma(t,T)F(t,T) dW_t + F(t,T)(\exp(e^{-a(T-t)}dJ_t)-1) + F(t,T)\mu_J(t,T)dt$$
I do not quite understand how this can be written in the integral form. i.e.
$$F(s,T) = F(0,T) + \int^s_0\sigma(t,T)F(t,T) dW_t + \int^s_0 F(t,T)\mu_J(t,T)dt + \cdots$$
I dont know what "$\cdots$" should be.
## Answer by Quantuple (score 3, accepted)
https://quant.stackexchange.com/a/31253
$\require{cancel}$
Consider the following SDE $$ \frac{dF(t,T)}{F(t,T)} = \sigma(t,T) dW_t + (\exp(e^{-a(T-t)}dN_t)-1) + \mu_J(t,T)dt $$ where $N_t$ figures a standard Poisson process, supposedly independent of the standard Brownian motion $W_t$.
This SDE should be interpreted by looking at $N_t$ as what it is, namely a random counting process with, intuitively, $dN_t$ zero everywhere except at (random) jump dates where it is equal to +1. Knowing this allows you to rewrite the jump term as: $$ \exp(e^{-a(T-t)}dN_t)-1 = \left(\exp(e^{-a(T-t)})-1\right)dN_t $$ which is better notational practice (the -1 which was standing out alone on the RHS was a bit weird), so that we have $$ \frac{dF(t,T)}{F(t,T)} = \mu_J(t,T)dt + \sigma(t,T) dW_t + \left(\exp(e^{-a(T-t)})-1\right)dN_t \tag{0} $$
Now, as hinted at by @Kiwiakos, letting $$ f(t,T) = \ln F(t,T) $$ and applying Itô's lemma for semi-martingales with jumps yields: $$ df(t,T) = \underbrace{\left(\mu_J(t,T)-\frac{\sigma(t,T)^2}{2}\right) dt + \sigma(t,T) dW_t}_{\text{diffusion part}} + \underbrace{(f(t,T)-f(t^-,T))dN_t}_{\text{jump part}} \tag{1} $$
The original SDE $(0)$ then tells us that, at a jump date $t$: $$ \underbrace{\frac{F(t,T) - F(t^-,T)}{F(t^-,T)}}_{dF(t,T)/F(t,T)} = \underbrace{0}_{\text{continuous part (does not jump)}} + \underbrace{ \exp(e^{-a(T-t)})-1}_{\text{non continuous part ($dN_t=1$)}} $$ or equivalently: $$ F(t,T) \cancel{- F(t^-,T)} = \exp(e^{-a(T-t)}) F(t^-,T) \cancel{- F(t^-,T)}$$ showing that at a jump date $$ f(t,T) - f(t^-,T) = \ln(F(t,T)/F(t^-,T)) = e^{-a(T-t)}$$ Hence the equivalent expression of the Itô lemma applied to $f(t,T)$ $$df(t,T) = \left(\mu_J(t,T)-\frac{\sigma(t,T)^2}{2}\right) dt + \sigma(t,T) dW_t + e^{-a(T-t)}dN_t \tag{2} $$
which can easily be integrated.
## Answer by Kiwiakos (score 1)
https://quant.stackexchange.com/a/31244
Looks like jump diffusion. You can take $f(t,T)=\log F(t,T)$, apply Ito's formula for jump diffusions and take it from there. I cannot see how taking integral directly can lead you anywhere.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.