Consider some model $$ dX_t = \mu d t + \sigma dW_t $$ where $\mu, \sigma$ are some constants. Now let $f \in C^{1,2}$ and consider $$ Y_t = f(t,X_t). $$ Say we (informally) consider a second order Taylor expansion then $$ Y_{t+h} - Y_{t} \approx f_t(t,X_t)\cdot h + f_x(t,X_t) (X_{t+h}-X_t) + 1/2 f_{xx} \cdot (X_{t+h}-X_t ) ^2 $$ which can be shown to converge (formally or informally) to the approximation resulting from applying Ito's formula and approximating the integrals by its left endpoint times the increment.

Now say I introduce jumps to $X$ i.e. say add a Compound Poisson process with normal jumps. I know how the Ito formula is changed to account for jumps, but have not studied noncontinuous processes much. Can the "Taylor approximation" intuition be maintained in some way in the non-continuous case? I.e. the added terms be motivated in a similar way.


Yes, Itô's formula for jump processes can be proved using Taylor's formula. For simplicity of notation, I'll just discuss that $f$ does not depend on the time $t$, i.e. $Y_t := f(X_t)$.

By Taylor's formula, we have

$$f(X_{t+h})-f(X_t) = f'(X_t) (X_{t+h}-X_t) + \frac{1}{2} f''(X_t) (X_{t+h}-X_t)^2 + R(X_t,X_{t+h})$$


$$R(x,y) := \left( \int_0^1 (1-\theta) f''(x+\theta (y-x)) \, d\theta \right) (y-x)^2 - \frac{1}{2} f''(x) (y-x)^2$$

denotes the remainder term. If $(X_t)_{t \geq 0}$ has continuous sample paths, then it is not difficult to see that the remainder term $R$ can be omitted, i.e.

$$f(X_{t+h}-f(X_t) \approx f'(X_t) (X_{t+h}-X_t) + \frac{1}{2} f''(X_t) (X_{t+h}-X_t)^2.$$

However, if $(X_t)_{t \geq 0}$ has jumps, this is no longer true. The reason is the following: If we let $h \to 0$, then

$$\begin{align*} R(X_t,X_{t+h}) &\xrightarrow[]{} \left( \int_0^1 (1-\theta) f''(X_t+\theta \Delta X_t) \, d \theta \right) (\Delta X_t)^2 - \frac{1}{2} f''(X_t) (\Delta X_t)^2 \\ &\stackrel{\text{Taylor}}{=} f(X_t)-f(X_{t-}) - f'(X_{t-}) \Delta X_t - \frac{1}{2} f''(X_t) (\Delta X_t)^2 \end{align*}$$

which means that we cannot expect that the remainder term vanishes as $h \to 0$. Consequently, we get an additional term in Itô's formula which is given by

$$\sum_{\Delta X_t \neq 0} \big[ f(X_t)-f(X_{t-}) - f'(X_{t-}) \Delta X_t - \frac{1}{2} f''(X_t) (\Delta X_t)^2 \big].$$


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