Let $X$ be geometric random variable with parameter $p$.
How to prove that:
(1) $E[X-1|X>1] = E[X]$
(2) $E[X^2|X>1] = E[(X+1)^2]$
Author explains the fact below and states it is used to prove (1) $$P(X-1=k|X>k) = P(X=k).$$
I understood how the fact is true but could not understand how it is used to derive (1)
(2) was stated without explanation. Could someone help in deriving (1) and (2)?
Firstly, I think your equation $P(X-1=k\mid X\gt k) = P(X=k)$ is supposed to be (proof below): $$P(X-1=k\mid X\gt 1) = P(X=k).\qquad\qquad (*)$$
(1)
\begin{eqnarray*} E(X-1\mid X\gt 1) &=& \sum_{k=1}^{\infty}{kP(X-1=k\mid X\gt 1)} \\ &=& \sum_{k=1}^{\infty}{kP(X=k)} \qquad\qquad\text{using $(*)$} \\ &=& E(X). \end{eqnarray*}
(2)
\begin{eqnarray*} E(X^2\mid X\gt 1) &=& \sum_{k=1}^{\infty}{k^2P(X=k\mid X\gt 1)} \\ &=& \sum_{k=1}^{\infty}{k^2P(X=k-1)} \qquad\qquad\text{using $(*)$ with $k-1$ instead of $k$} \\ &=& \sum_{k=1}^{\infty}{(k+1)^2P(X=k)} \qquad\qquad\text{re-index the sum using $P(X=0)=0$} \\ &=& E((X+1)^2). \end{eqnarray*}
Note the different mechanism in first line of the two methods. In (1) we treat $X-1$ as a random variable in itself, whereas in (2) we treat $X^2$ as a function of the random variable $X$.
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Claim: $P(X-1=k\mid X\gt 1) = P(X=k)$.
Firstly, for $k=0$ it's clear that both sides are $0$. So now assume $k\gt 0$. \begin{eqnarray*} P(X-1=k\mid X\gt 1) &=& P(X-1=k\cap X\gt 1) / P(X\gt 1) \\ &=& P(X=k+1\cap X\gt 1) / P(X\gt 1) \\ &=& P(X=k+1) / (1-P(X=1)) \\ &=& (1-p)^kp / (1-p) \\ &=& (1-p)^{k-1}p \\ &=& P(X=k). \end{eqnarray*}