Martingales – Problem (30/365)

Let (\theta_k, \mathcal{D}_k) and (\phi_k, \mathcal{D}_k) be two martingales, \theta_1 = \phi_1 = 0. Show that

\displaystyle  E\theta\phi = \sum_{k=2}^{n}E(\theta_{k}-\theta_{k-1})(\phi_{k}-\phi_{k-1})

Proceed as follows.

\displaystyle  \sum_{k=2}^{n}E(\theta_{k}-\theta_{k-1})(\phi_{k}-\phi_{k-1}) \\ = E\sum_{k=2}^{n}E\left[(\theta_{k}-\theta_{k-1})(\phi_{k}-\phi_{k-1})|\mathcal{D}_{k-1}\right]\mbox{ by total probability formula} \\ = E\sum_{k=2}^{n}E(\theta_{k}\phi_{k}|\mathcal{D}_{k-1})-E(\theta_{k}\phi_{k-1}|\mathcal{D}_{k-1})-E(\theta_{k-1}\phi_{k}|\mathcal{D}_{k-1})+E(\theta_{k-1}\phi_{k-1}|\mathcal{D}_{k-1}) \\ = E\sum_{k=2}^{n}E(\theta_{k}\phi_{k}|\mathcal{D}_{k-1})-\phi_{k-1}E(\theta_{k}|\mathcal{D}_{k-1})-\theta_{k-1}E(\phi_{k}|\mathcal{D}_{k-1})+E(\theta_{k-1}\phi_{k-1}|\mathcal{D}_{k-1}) \\ \mbox{ because }\phi_{k-1},\theta_{k-1}\mbox{ are }\mathcal{D}_{k-1}\mbox{-measurable} \\ = E\sum_{k=2}^{n}E(\theta_{k}\phi_{k}|\mathcal{D}_{k-1})-2\phi_{k-1}\theta_{k-1}+E(\theta_{k-1}\phi_{k-1}|\mathcal{D}_{k-1})\mbox{ by definition of martingale} \\ = E\sum_{k=2}^{n}E(\theta_{k}\phi_{k}|\mathcal{D}_{k-1})-\phi_{k-1}\theta_{k-1}\mbox{ because }\theta_{k-1}\phi_{k-1}\mbox{ is }\mathcal{D}_{k-1}\mbox{-measurable} \\ = E\sum_{k=2}^{n}E(\theta_{k}\phi_{k}|\mathcal{D}_{k-1})-E(\phi_{k-1}\theta_{k-1}|\mathcal{D}_{k-1}) \\ = E\theta_{n}\phi_{n}-E\phi_{1}\theta_{1} \\ = E\theta_{n}\phi_{n}\mbox{ since }\theta_{1}=\phi_{1}=0

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