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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.

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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.

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    $\begingroup$ A nice solution +1 :) $\endgroup$ – user16651 Nov 29 '16 at 16:47
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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.

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  • $\begingroup$ are you saying this sde cannot be written in the integral form? $\endgroup$ – Lost1 Nov 29 '16 at 5:52
  • $\begingroup$ No, I am not saying that. Of course you can write $...=\int_0^s F(t,T) [\exp(a(T-t)dJ_t)-1]$. I am saying that this does not appear to lead to a solution. $\endgroup$ – Kiwiakos Nov 29 '16 at 7:30
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    $\begingroup$ Well can you? The -1 standing alone is a bit weird IMHO $\endgroup$ – Quantuple Nov 29 '16 at 9:16
  • $\begingroup$ What's the difference? If $dJ=0$ then the term is zero; if $dJ=1$ then the term is $\exp[e^{-a(T-t)}-1]$. Therefore it's the same as $\exp[...]dJ$. $\endgroup$ – Kiwiakos Nov 29 '16 at 12:43
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    $\begingroup$ I agree that the interpretation is the same, but I meant to point out that notation is not a good practice IMHO as it can make one think he/she can write $\int_0^s F(t,T) [\exp(a(T-t)dJ_t)-1] = \int_0^s F(t,T) \exp(a(T-t)dJ_t - \int_0^s F(t,T) 1$ but unfortunately, this does not mean anything. $\endgroup$ – Quantuple Nov 29 '16 at 15:04

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