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Let $\{X_i\}$ be a sample from a normal $N(\theta,1)$, where $\theta \in \Bbb Z$. Show the estimator $$ T=\left\lfloor \bar X _ n \right \rfloor, $$ is not consistent for $\theta$, with $\bar X_n$ the sample mean of an $n$-sample.

Let $\epsilon >0$, I should show $$ \lim_{n \to \infty}P\left( |T-\theta| > \epsilon \right) = \lim_{n \to \infty} P\left( \left|\left\lfloor \bar X _ n \right \rfloor - \theta \right| > \epsilon \right) \ne 0, $$ but the integer part function confuses me.
I don't see the distribution of the random variable in the middle term.

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Hint: If $\bar X_n<\theta$, then $\left\lfloor \bar X_n \right \rfloor\leq \theta-1$. What is $P(\bar X_n<\theta)$?

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  • $\begingroup$ \begin{align} P\left( \left \lfloor \bar X_n \right \rfloor - \theta < -1/2 \right) &= P\left( \left \lfloor \bar X_n \right \rfloor < \theta -1/2 \right) P\left( \bar X_n < \theta \right) \\ &+P\left( \left \lfloor \bar X_n \right \rfloor < \theta -1/2 \right) P\left( \bar X_n \ge \theta \right) \\ &\ge 1 \cdot 1/2 + 0 \cdot 1/2 \\ &= 1/2. \end{align} Is it like that? $\endgroup$
    – user14108
    Commented Feb 23, 2013 at 0:27
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    $\begingroup$ @NicolasEssis-Breton I don't follow your calculations at all. Using my hint, get some idea about $P\left(| \left \lfloor \bar X_n \right \rfloor - \theta| \geq 1\right)$. $\endgroup$
    – user940
    Commented Feb 23, 2013 at 0:31

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