Quadratic Variation of Brownian Motion with a Poisson Process
Summary
The note works through the quadratic variation of a process formed by a Brownian stochastic integral with time-varying volatility and an independent unit-jump Poisson process. The continuous component contributes the integral of its squared diffusion coefficient, while the jump component contributes the sum of squared jump sizes. Since each Poisson jump has size one, its quadratic variation increases by one at every jump and matches the counting process.
The answer justifies this result from the partition definition of quadratic variation: refining partitions isolates the jumps, whose squared increments equal one. The cross variation between the continuous Brownian part and the jump process vanishes, yielding the sum of the two components' quadratic variations. This example assumes a standard Poisson process with unit jumps; processes with different jump sizes would contribute their squared jump sizes instead.
Key ideas
- The quadratic variation of a continuous Brownian integral is the integral of its squared volatility coefficient.
- A unit-jump Poisson process has quadratic variation equal to its counting process.
- Quadratic variation accumulates squared jump sizes at jump times.
- The mixed process's quadratic variation combines its continuous and jump contributions.
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# Quadratic Variation Of Mixed Brownian Motion and Poisson Process
# Quadratic Variation Of Mixed Brownian Motion and Poisson Process
I am trying to solve this problem where we're asked to compute the quadratic variation of a process. I assume that it is necessary to apply Ito's formula but not sure how to get the right solution. Furthermore, I'm also not sure about how to apply Ito's formula to a function that includes integrals like this one. I am familiar with the basic Ito Formula and know how to apply it to simpler functions.
Let N be a (P,F)-Poisson process with parameter $\lambda$ > 0 and define the process: X = $(X)_t$
$$ X_t = 2 + \int_{0}^{t} \sqrt{s} dW_s + N_t $$
Compute $[X]_t$
is this attempt correct?
$$ X_t = 2 + \int_{0}^{t} \sqrt{s} dW_s + N_t $$
which we can write in differential form as:
$$ dX_t = \sqrt{t}dW_t + dN_t $$
then the quadratic variation is given as:
$$ d[X_t] = dX_t * dX_t = (\sqrt{t}dW_t + dN_t) * (\sqrt{t}dW_t + dN_t) = tdt + dNt $$
I assumed that the cross terms cancel out. Then we can then rewrite it in integral form to get $[X_t]$
$$ [X_t] = \int_{0}^{t} t dt + \int_{0}^{t} dN_t = \frac{t^2}{2} + N_t $$
If this attempt is correct, I am not quite sure why $dN_t * dN_t = dN_t$ would be true in the quadratic covariation step.
## Answer by Daneel Olivaw (score 6, accepted)
https://quant.stackexchange.com/a/75578
Your attempt is correct.
The quadratic variation for a Poisson process is: $$[N]_t=\lim_{\sup(t_{i+1}-t_i)\rightarrow0}\sum_{i:t_i\leq t}(N_{t_{i+1}}-N_{t_i})^2\tag{1}$$ for some partition $\Pi(t)=0\leq t_0\leq\dots\leq t_n\leq t$ of the segment $[0,t]$. Simply pick the partition such that the jump times $\tau_1\leq\dots\leq\tau_k$ of $N$ between $0$ and $t$ are included in $\Pi(t)$ then given $N_{\tau_j}-N_{\tau_j^-}=\Delta N_{\tau_j}$: $$[N]_t=\sum_{j:\tau_j\leq t}\Delta N_{\tau_j}^2$$ But $\Delta N_{\tau_j}$ is always equal to 1, and so is $\Delta N_{\tau_j}^2$, therefore the quadratic variation of the Poisson process jumps by $1$ whenever $N$ jumps. Consequently: $$\textrm{d}[N]_t=\textrm{d} N_t$$
Addendum. To justify picking the partition $\Pi(t)$ such that it contains the jump times $\tau_1\leq\dots\leq\tau_k$, consider the definition $(1)$ again. For each $i$, either there is no $j\in[1,k]$ such that $\tau_j\in[t_i,t_{i+1})$ in which case $\smash{\lim_{\sup(t_{i+1}-t_i)\rightarrow0}(N_{t_{i+1}}-N_{t_i})^2=0}$; else assuming without loss of generality that there is a single jump in the interval $[t_i,t_{i+1})$: \begin{align} N_{t_{i+1}}-N_{t_i} &=(N_{t_{i+1}}-N_{\tau_j})+(N_{\tau_j}-N_{t_i})\\ &=(N_{\tau_j}-N_{\tau_j})+(N_{\tau_j}-N_{\tau_{j-1}})\\ &=\Delta N_{\tau_j} \end{align} from which it entails the original argument is valid $-$ the argument extends to an arbitrary number of jumps within $[t_i,t_{i+1})$ at the cost of more involved notation.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.