Solution: Doob decomposition of a partial sum

Eksempel

Let \(X_0 = 0\) and \(X_n = U_1 + \ldots + U_n\) for \(n = 1, 2,\ldots\), where \(U_1, U_2,\ldots\) are independent and identically distributed stochastic variables with \(E[U_i ] = \mu\). We are to find the Doob decomposition of the process \(X\).

Solution

From \((2.16)\) in ABG we have that the Doob decomposition is given by \[X_n = E[X_n | \mathcal{F}_{n−1}] + \Delta M_n,\] where \[\Delta M_n = X_n - E[X_n | \mathcal{F}_{n−1}] .\] Expressing the \(X_n\) using the \(U_i\), and noting that \(U_i\) is known given \(\mathcal{F}_{n-1}\) when \(i \leq n-1\) and that \(U_n\) is independent of \(\mathcal{F}_{n-1}\) we get \begin{align*} E[X_n | \mathcal{F}_{n−1}] &= E\left[\sum_{i=1}^n U_i | \mathcal{F}_{n−1}\right] \\ &= E\left[\sum_{i=1}^{n-1} U_i | \mathcal{F}_{n−1}\right] + E\left[U_n | \mathcal{F}_{n−1}\right] \\ &= \sum_{i=1}^{n-1} U_i + E\left[U_n\right] \\ &= X_{n-1} + \mu.\end{align*} Hence the predictable (observe that it only depends on a prior value) part of the Doob decomposition is \[\underline{\underline{E[X_n | \mathcal{F}_{n−1}] = X_{n-1} + \mu.}}\] Using this further we get \begin{align*} \Delta M_n &= X_n - E[X_n | \mathcal{F}_{n−1}] \\ &= X_n - (X_{n-1} + \mu)\\ &= X_n -X_{n-1} - \mu \\ &= U_n - \mu.\end{align*} Thus the innovation part of the Doob decomposition is \[\underline{\underline{\Delta M_n = U_n - \mu.}}\]