# Crossvariation computation

Suppose $$M$$ and $$N$$ are continuous martingales, $$K$$ and $$H$$ are processes that can be written in the form: $$K_s = \sum_{i=0}^p Y_{a_i}\mathbb{1}_{(a_i,a_{i+1}]}(s),\quad H_s = \sum_{j=0}^q X_{a_j}\mathbb{1}_{(a_j,a_{j+1}]}(s)$$ with each $$Y_{a_i}$$ $$\mathcal{F}_{a_i}$$-measurable. For such processes we define: $$\int_0^t K_s dM_s = \sum_{i=0}^p Y_{a_i}(M_{t \land a_{i+1}}-M_{t \land a_i})$$ I'm trying to compute the cross-variation $$\left\langle \int_0^{.} K_s dM_s,\int_0^{.} H_s dN_s \right\rangle_t$$ but I get lost in the computation of: $$\left\langle \sum_{i=0}^p Y_{a_i}(M_{\cdot \land a_{i+1}} - M_{\cdot \land a_{i}}) ,\sum_{j=0}^q X_{a_j}(N_{\cdot \land a_{j+1}} - N_{\cdot \land a_{j}})\right\rangle_t$$ obtaining something like: $$\sum_{i=0}^p \sum_{j=0}^q Y_{a_i}X_{a_j}\langle M_{\cdot \land a_{i+1}} - M_{\cdot \land a_{i}}, N_{\cdot \land a_{j+1}} - N_{\cdot \land a_{j}} \rangle_t$$ What am I doing wrong?

P.S. The result should be: $$\int_0^t K_s H_s d \langle M, N \rangle_s$$

• Somewhere you dropped the $t\wedge$ from $M_{t\wedge\, a_i}$ etc. You should figure out what $\langle M_{\,\cdot\,\wedge\, a_i}-M_{\,\cdot\,\wedge \,a_{i+1}},N_{\,\cdot\,\wedge\, a_i}-N_{\,\cdot\,\wedge\, a_{i+1}}\rangle_t$ is. Jul 26, 2021 at 9:02
• @KurtG. That's indeed my difficulty. I notice you used only the $i$ index, but a priori the $a_i$ and $a_j$ should be allowed to be different, right? Jul 26, 2021 at 10:10

Your question is a duplicate. In the answer there your approach is suggested as an alternative but the details are left as an exercise. Does the following help? The product of your stochastic integrals is $$\sum_{i=0}^p\sum_{j=0}^q Y_{a_i}X_{a_j} (M_{t \wedge a_{i+1}} - M_{t \wedge a_{i}}) (N_{t \wedge a_{j+1}} - N_{t \wedge a_{j}})\,.$$ For each $$t$$ there is a unique pair $$(k,l)$$ such that this is $$\underbrace{\sum_{i=0}^{k-1} \sum_{j=0}^{l-1} Y_{a_i}X_{a_j} (M_{a_{i+1}} - M_{a_i})(N_{a_{j+1}} - N_{a_j})}_{=:Z_{k,l}}+Y_{a_k}X_{a_l} (M_t - M_{a_k})(N_{t} - N_{a_l})\,.$$ This is clearly of the form $$Z_{k,l}+Y_{a_k}X_{a_l} (M_tN_t - M_{a_k}N_t - N_{a_l}M_t+M_{a_k}N_{a_l}).$$ The process with bounded variation that we have to subtract from this to make it a martingale is obviously $$Z_{k,l}+Y_{a_k}X_{a_l} (\langle M,N\rangle_t+M_{a_k}N_{a_l}).$$