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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: $$ \int_0^t K_s H_s d \langle M, N \rangle_s $$

Dalamar
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  • 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. – Kurt G. Jul 26 '21 at 09:02
  • @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? – Dalamar Jul 26 '21 at 10:10

1 Answers1

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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 $$ Z_{k,l}+Y_{a_k}X_{a_l} (\langle M,N\rangle_t+M_{a_k}N_{a_l}). $$

Kurt G.
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