I am having a hard time following this proof, maybe the solutions have jumped some steps but was wondering if someone could help me follow it.
The question:
let $X_{1},...,X_{n}$ be independent and identically distributed sequence of random variables from a population where $g_{\theta}(x) = 2^{{\theta}-x-1}$ where $x \in (\theta, \theta+1,\theta+2...)$
Show that the statistic $T_{2}(\vec{X}) = max (X_{1},X_{2})$ is not sufficient for $\theta_{0}$
We note that $X_{1},...,X_{n}$ are independent and that $T_{2}(\vec{X})$ only depends on $X_{1}$ and $X_{2}$. Therefore, the distribution of $X_{3}$ conditional on $T_{2}(\vec{X})$ is the same as $g_{\theta}$. Since the original $g_{\theta}$ depends on $\theta$ we conclude that the distribution of $X_{3}$ conditional on $T_{2}(\vec{X})$ also depends on $\theta$ and thus $T_{2}$ is not a sufficient statistic.
Proof that distribution depends on $\theta$
If $P(X_{3} = x) = 2^{{\theta}-x-1}$, which depends on $\theta$
$E[X_{3}] = \sum_{x=\theta}^{\inf}x2^{{\theta-x-1}}$ = $\sum_{x=0}^{\inf}(x+\theta)2^{{-x-1}} = \sum_{x=1}^{\inf}x2^{{-x-1}}+\sum_{x=1}^{\inf}\theta2^{-x}$ = $2^{-1}\sum_{x=1}^{\inf}x2^{-x} +\sum_{x=1}^{\inf}\theta2^{-x} = 2+\theta$
But I am unsure how they got the last line of this proof?
Was hoping someone could help fill in the blanks.
Many thanks.
It looks like a factor of $2^{-1}$ was dropped, either in the solution or your typing it out here. Maybe that's the trouble? Recall from calculus that for $|r| < 1$,
$$ \sum_{n=1}^{\infty}r^n = \frac{r}{1-r}\\ \sum_{n=1}^{\infty}nr^n = \frac{r}{(1-r)^2}\\ $$
So,
$$ \begin{align} E[X] &= \sum_{x = \theta}^{\infty}x2^{\theta-x-1}\\ &=\sum_{x=0}^{\infty}(x+\theta)2^{-x-1}\\ &=\sum_{x=0}^{\infty}x2^{-x-1} + \sum_{x=0}^{\infty}\theta2^{-x-1}\\ &=\sum_{x=1}^{\infty}x2^{-x-1} + \theta2^{-1}+\sum_{x=1}^{\infty}\theta2^{-x-1}\\ &=2^{-1}\sum_{x=1}^{\infty}x2^{-x} + \theta 2^{-1} + \theta2^{-1}\sum_{x=1}^{\infty}2^{-x}\\ &=2^{-1}\sum_{x=1}^{\infty}x \left( \frac{1}{2} \right) ^x + \theta 2^{-1} + \theta2^{-1}\sum_{x=1}^{\infty} \left( \frac{1}{2}\right)^{x}\\ &= 1 + \theta 2^{-1} + \theta 2^{-1}\\ & = 1 + \theta \end{align} $$