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:
$$ d[X_t] = dX_t * dX_t = (\sqrt{t}dW_t + dN_t) * (\sqrt{t}dW_t + dN_t) = tdt + dNt $$
which we can then write again 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. I assumed that the cross terms cancel out.
Best Answer
With a sequence of partitions $0=t^n_0<\dots< t^n_n=t$ and $\Pi_n:=\max\limits_{i=1,\dots, n}|t^n_i-t^n_{i-1}|$ we have by definition \begin{align} [W,N]_t&=\lim_{\scriptstyle n\to\infty\atop\scriptstyle\Pi_n\to\, 0}\sum_{i=1}^n\Big(W_{t^n_i}-W_{t^n_{i-1}}\Big)\Big(N_{t^n_i}-N_{t^n_{i-1}}\Big)\,. \end{align} Since $N_t$ is increasing and $W_t$ is continuous (therefore uniformly continuous on $[0,t]$), \begin{align} \Big|[W,N]_t\Big|&\le\lim_{\scriptstyle n\to\infty\atop\scriptstyle\Pi_n\to\, 0}\sum_{i=1}^n\Big|W_{t^n_i}-W_{t^n_{i-1}}\Big|\,\Big|N_{t^n_i}-N_{t^n_{i-1}}\Big|\\ &\le\lim_{\scriptstyle n\to\infty\atop\scriptstyle\Pi_n\to\, 0}\Bigg\{\underbrace{\max_{i=0,\dots,n}\Big|W_{t^n_i}-W_{t^n_{i-1}}\Big|}_{\textstyle\to 0}\cdot\underbrace{\sum_{i=1}^nN_{t^n_i}-N_{t^n_{i-1}}}_{\textstyle N_t}\Bigg\}=0\,. \end{align}
Edit: Your end result is correct but you could do a lot to improve the notation. I avoid physics notation such as $$ dX_t*dX_t $$ (not only because of that awful $*$) and prefer to calculate quadratic variations in integral form. The relations $$ [W]_t=t\,,\quad [N]_t=N_t\,,\quad [W,N]_t=0 $$ then lead to $$ [X]_t = \int_{0}^{t} \color{red}{s\,ds} + N_t=\frac{t^2}{2}+N_t\,. $$ From there you can go to its differential form $$ d[X]_t=t\,dt+dN_t $$ if you want.