Moment of Geometric Distribution

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

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



Answers 1


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

Sherwin Lott
Sherwin Lott
June 30, 2020 02:19 AM

Related Questions


Second Moment of shifted Geometric distribution

Updated July 01, 2020 04:20 AM


Statistical Test for a Poisson Distribution

Updated September 04, 2019 01:20 AM


Predict the daily usage of Bandwidth of a Network

Updated July 25, 2016 08:08 AM