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?


1 Answer 1


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

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

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

  • $\begingroup$ 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? $\endgroup$
    – Giogre
    Jul 13, 2021 at 17:22
  • $\begingroup$ @Giogre it’s because the quadratic variation, that is $[\cdot,\cdot]_t$, between a Brownian Motion and a jump process is null. $\endgroup$ Jul 13, 2021 at 17:33
  • $\begingroup$ 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. $\endgroup$
    – ir7
    Jul 13, 2021 at 17:40
  • $\begingroup$ 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? $\endgroup$
    – Giogre
    Jul 13, 2021 at 18:10
  • 1
    $\begingroup$ I included an elementary proof of that fact starting from the quadratic covariation definition here on the Stack. $\endgroup$
    – ir7
    Jul 13, 2021 at 18:19

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