# Moment of Geometric Distribution

by Ab2020   Last Updated June 30, 2020 02:20 AM

Suppose $$Z\sim$$Geo(p). Explain that $$E[Z^2] = p + (1-p) E[(1+Z)^2]$$

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\begin{align} E[Z^2] &= P(Z=1)E[Z^2|Z=1] + P(Z>1)E[Z^{2} | Z > 1]\\ &= p + (1-p)E[(1+Z)^2] \end{align} Note that $$E[Z^{2} | Z > 1] = E[(1+Z)^2]$$ because future coin flips are independent of earlier ones.