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We have

\begin{aligned} E(Z_1) = A \\ \Pr \{ Z_2 = Z_1 + 1 \} = \frac 1 2 \\ \Pr \{ Z_2 = Z_1 - 1 \} = \frac 1 2 \end{aligned}

Are these conditions enough to get $E(Z_2)$?

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up vote 3 down vote accepted

Yes, because $\newcommand{\E}{\mathbf{E}}$ $$ \E[Z_2] = \E[Z_1] + \E[Z_2-Z_1] $$ holds for all random variables $Z_1$ and $Z_2$ (and you can calculate each term on the right). In other words, linearity of expectations can be applied in all cases, however correlated the random variables may be.

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