# Second variation of a Brownian motion under jump-diffusion process

I am trying to solve exercise 15.3 from the book The concepts and practice of mathematical finance where it is asked

Suppose the $$\log S_t$$ follows a Brownian motion over the period $$[0, 1]$$ except at time $$0.5$$ where it jumps by $$x$$. What are the first and second variations of $$\log S_t$$ over the period $$[0, 1]$$.

The first variation is easily determined to be $$\infty$$, as in a continuous Brownian motion.

A continuous Brownian motion should also have its second variation equal to $$T$$, so here - not considering the jump - it would be equal to $$T = 1$$. But unlike the first variation, the second one is a finite value, so should be susceptible to the presence of the jump.

Indeed the solution reported in the book states the second variation to be equal to $$1.25$$.

Where does this result comes from?

$$X_t = B_t 1_{t<0.5} + (x+ B_t) 1_{t\geq 0.5} = B_t + x1_{t\geq 0.5}$$

$$[X, X]_t = [B, B]_t + x^2 1_{t\geq 0.5} = t+ x^2 1_{t\geq 0.5}$$

(the author probably intended to use $$0.5$$ as jump size too)

• Nice. The double product $B_t x$ in the last expression is tossed out because the mean of $B_t$ is assumed $= 0$, am I right? Jul 13 at 17:22
• @Giogre it’s because the quadratic variation, that is $[\cdot,\cdot]_t$, between a Brownian Motion and a jump process is null. Jul 13 at 17:33
• Yes, $B_t$ is a continous process while $Y_t = x1_{t\leq 0.5}$ is a finite variation process, so their quadratic covariation is $0$. See Lemma 3 here.
– ir7
Jul 13 at 17:40
• I see the link provided. I imagined the Concepts book was self-contained, not requiring more stochastic calculus than that already in the book. Basically $x 1_{t \geq 0.5}$ is a step function hence its variation should be zero. Is this a good intuitive explanation? Then why should the quadratic covariation be $0$ only if the other process is continuous? What changes if it were another step function, for instance? Jul 13 at 18:10
• I included an elementary proof of that fact starting from the quadratic covariation definition here on the Stack.
– ir7
Jul 13 at 18:19