Why $E[X|X>1] = E[X] + 1$ for geometric random variable? For a geometric random variable X, $P_X(k) = (1-p)^{k-1}p$, then
$$E[X] = \sum_{k=1}^{\infty}k(1-p)^{k-1}p\;.$$ That is ($q = 1 -p$)
$$E[X] = p + 2qp + 3q^2p + 4q^3p + \dotso $$
But 
$$P_{X|X>1}(k) = \begin{cases}(1-p)^{k-2}p\text{ if }k > 1\\
0\text{ if }k = 1\end{cases}$$
$$E[X|X>1] = \sum_{k=2}^{\infty}k(1-p)^{k-2}p\;,$$ that is
$$E[X|X>1] + 1 = 1 + 2p + 3qp +  4q^2p + \dotso$$
So how could $E[X|X>1] = E[X] + 1$?
Thanks.
 A: Since $P_X(k)$ is normalized, we have 
$$\sum_{k=1}^\infty(1-p)^{k-1}p=\sum_{k=2}^\infty(1-p)^{k-2}p=1\;.$$
Thus
$$
\begin{eqnarray}
E[X|X>1]
&=&
\sum_{k=2}^{\infty}k(1-p)^{k-2}p
\\
&=&
\sum_{k=2}^{\infty}(k-1)(1-p)^{k-2}p+\sum_{k=2}^{\infty}(1-p)^{k-2}p
\\
&=&
\sum_{k=1}^{\infty}k(1-p)^{k-1}p+1
\\
&=&
E[X] + 1
\end{eqnarray}
$$
All this is really saying is that since the conditional probability for $k+1$ is the same as the unconditional probability for $k$, the conditional expectation value of $k$ must be the unconditional expectation value of $k+1$.
A: A coin has probability $p$ of landing heads, and $q=1-p$ of landing tails. Assume that $p\ne 0$.
Let $X$ be the total number of tosses until you get a head.  Then $X$ has precisely the geometric distribution that you described. One can, as you did, get an expression for $E(X)$ as an infinite series.  In fact, it turns out $E(X)=1/p$.  But we need neither the series nor its sum to prove the result that is asked for.
Suppose that we are given that $X>1$.  This means that our first toss was a tail. Let $Y$ be the additional number of tosses that we must wait for a head.  The coin does not remember that the first toss was a tail, so $Y$ has the same distribution, and therefore the same mean, as $X$.  In symbols, $E(Y)=E(X)$.
But the total number of tosses, given that $X>1$, is $1+Y$.  The $1$ is for the "wasted" first toss. Thus
$$E(X|X>1)=E(1+Y)=1+E(Y)=1+E(X).$$
Comment: If you prove the result using the infinite series, you know that the result is true.  If you do it more conceptually, you know why the result is true. 
A: The "memorylessness" of the geometric distribution implies that the conditional probability distribution of $X$ given that $X\ge\text{any particular integer}$ is the same as the probability distribution of $X+\text{that same integer}$.
