Wald first equation under integrable stopping

Pointwise, ${\sum }_{k=1}^{\tau }{X}_k={\sum }_{k\ge 1}{X}_k{1}_{\{\tau \ge k\}},\tau ={\sum }_{k\ge 1}{1}_{\{\tau \ge k\}}\mathrm{.}$ The first identity is understood through finite partial sums; the integrability calculation below shows that $\tau =\mathrm{\infty }$ is null.

By F1, F2, F3, $\mathbb{E}[\mathrm{\mid }{X}_k\mathrm{\mid }{1}_{\{\tau \ge k\}}]=\mathbb{E}\mathrm{\mid }{X}_1\mathrm{\mid }\mathbb{P}(\tau \ge k)\mathrm{.}$ MCT applied to the nonnegative partial sums and to the second identity in step 1.1 gives $\mathbb{E}{\sum }_{k\ge 1}\mathrm{\mid }{X}_k\mathrm{\mid }{1}_{\{\tau \ge k\}}=\mathbb{E}\mathrm{\mid }{X}_1\mathrm{\mid }{\sum }_{k\ge 1}\mathbb{P}(\tau \ge k)=\mathbb{E}\mathrm{\mid }{X}_1\mathrm{\mid }\mathbb{E}\tau <\mathrm{\infty }\mathrm{.}$ Thus the stopped series is absolutely integrable and $\tau <\mathrm{\infty }$ almost surely.

The signed partial sums are dominated by the integrable absolute series in step 2.1. DCT and finite linearity therefore give $\mathbb{E}{\sum }_{k=1}^{\tau }{X}_k={\sum }_{k\ge 1}\mathbb{E}[X_k{1}_{\{\tau \ge k\}}]={\sum }_{k\ge 1}\mu \mathbb{P}(\tau \ge k)=\mu \mathbb{E}\tau ,$ where F3 gives the middle factorization. This argument is choice-free: the given iid sequence and stopping time supply all indexed objects, and no conditional-expectation version is chosen.