Quadratic Variation of Brownian Motion with a Single Jump

Summary

The exchange analyzes the second variation, or quadratic variation, of a process whose path follows Brownian motion and includes a single jump at the midpoint of the observation interval. Brownian motion contributes quadratic variation equal to elapsed time. A jump contributes the square of its size, so the process has quadratic variation equal to time plus that squared jump once the jump has occurred.

This explains why the terminal value can exceed the continuous Brownian contribution. The cited exercise gives a terminal value of 1.25; the answer infers that this corresponds to a jump size of 0.5, apparently treating the stated midpoint time as the jump magnitude. The first variation is described in the question as infinite for the Brownian component, but the answer focuses on quadratic variation. The numerical result therefore depends on the inferred jump size; the jump time alone does not determine the additional variation.

Key ideas

• Brownian motion accumulates quadratic variation at a rate of one per unit of time.
• A discrete jump adds the square of its magnitude to quadratic variation.
• After a jump, total quadratic variation combines elapsed time and the squared jump size.
• The exercise's terminal value is consistent with a jump magnitude of 0.5, as the answer infers.
• The jump time by itself does not specify the jump's contribution.

Tags

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


# 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?

## Answer by ir7 (score 3, accepted)

https://quant.stackexchange.com/a/66008

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

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

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

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