Martingales – Problem (28/365)

Let the random variables \eta_1, \dots, \eta_k satisfy E(\eta_k | \eta_1, \dots, \eta_{k-1})=0. Show that the sequence \theta = (\theta_k)_{1\le k \le n} with \theta_1 = \eta_1 and

\displaystyle  \theta_{k+1} = \sum_{i=1}^k \eta_{i+1}f_i(\eta_1, \dots, \eta_i)

where f_i are given functions, is a martingale.

Once again, we let the sequence of decompositions be \mathcal{D}_k = \mathcal{D}_{\eta_1, \dots, \eta_k}. Then \eta_k is \mathcal{D}_k-measurable because (1) f_i is D_{k \ge i}-measurable, (2) f_{i+1} f_i is \mathcal{D}_{i+1}-measurable, and (3) f_i + f_j is \mathcal{D}_i-measurable if i \ge j.

Next, we show that E(\theta_{k+1} | \mathcal{D}_k) = \theta_k

\displaystyle  E(\theta_{k+1}|\mathcal{D}_{k})	\\ = E\left[\sum_{i=1}^{k}\eta_{i+1}f_{i}(\eta_{1},\dots,\eta_{i})|\mathcal{D}_{k}\right] \\ = \sum_{i=1}^{k}E\left[\eta_{i+1}f_{i}(\eta_{1},\dots,\eta_{i})|\mathcal{D}_{k}\right] \\ = \sum_{i=1}^{k}E\left[\eta_{i+1}|\mathcal{D}_{k}\right]f_{i}(\eta_{1},\dots,\eta_{i})\mbox{ since }f_{i}(\eta_{1},\dots,\eta_{i})\mbox{ is }\mathcal{D}_{k\ge i}\mbox{-measurable} \\ = E\left[\eta_{k+1}|\mathcal{D}_{k}\right]f_{i}(\eta_{1},\dots,\eta_{k})+\sum_{i=1}^{k-1}E\left[\eta_{i}|\mathcal{D}_{k}\right]f_{i}(\eta_{1},\dots,\eta_{i}) \\ = \sum_{i=1}^{k-1}E\left[\eta_{i}|\mathcal{D}_{k}\right]f_{i}(\eta_{1},\dots,\eta_{i})\mbox{ since we are given }E\left[n_{k}|\eta_{1},\dots,\eta_{k-1}\right]=0 \\ = \sum_{i=1}^{k-1}\eta_{i+1}f_{i}(\eta_{1},\dots,\eta_{i})\mbox{ since }\eta_{i}\mbox{ is }\mathcal{D}_{k\ge i}\mbox{-measurable} \\ = \theta_{k}

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