The expected value of a fractional part of a random variable Let $X\sim N(\mu, 1)$ where it is known that $ \mu$ Is an integer.
Define an estimator $\delta$ by taking the sample mean and rounding it to the nearest integer.
My aim is to show that given the restricted parameter space, the estimator is unbiased.
My attempt:
By expressing the random variable in terms of indicator functions floors and fractional parts, we can write the expected value of $\delta $ as
$$E \delta = E floor(X_n) + P(fractional (X_n) >0.5)$$
Hence if I can show that 
$$ E fractional(X_n) = P(fractional (X_n) >0.5)$$ then the result follows. Intuitively I think this makes sense, but I’m not sure how to proceed further. So far I have not used the fact the parameter space is restricted, but I think it must play a role in the proof some how 
 A: Let's write $\{x\}$ to represent the fractional part of $x$
Given the value of $\mu$, we can let $Y=X -\mu$.  We then have:


*

*$Y \sim N(0,1)$ since we have subtracted the mean of $X$

*$E[\{Y\}] = E[\{X\}]$ and $P\left(\{Y\} > \frac12\right) = P\left(\{X\} > \frac12\right)$ since we have subtracted an integer

*$P\left(\{Y\} > \frac12 \mid Y >0\right)=P\left(\{Y\} \le \frac12 \mid Y \le 0\right)$ and  $P\left(\{Y\} \le  \frac12 \mid Y >0\right)=P\left(\{Y\} \gt \frac12 \mid Y \le 0\right)$ by symmetry of the standard normal 

*$P\left(\{Y\} > \frac12\right) = \frac12P(\{Y\} > \frac12 \mid Y >0) +\frac12P(\{Y\} > \frac12 \mid Y \le 0) \\ = \frac12P(\{Y\} > \frac12 \mid Y >0) +\frac12P(\{Y\} \le \frac12 \mid Y > 0) = \frac12$

*$E[\{Y\} \mid Y >0] = E[1-\{Y\} \mid Y \le 0] $ so $E[\{Y\} \mid Y >0] + E[\{Y\} \mid Y \le 0]=1$

*$E[\{Y\}] = \frac12E[\{Y\} \mid Y >0] + \frac12E[\{Y\} \mid Y \le 0] =\frac12$
and this leads to $E[\{X\}]=\frac12= P\left(\{X\} > \frac12\right)$ for given $\mu$ and so for all $\mu$ as requested
Similar manipulations lead to seeing that $\{Y\}$ and $\{X\}$ are uniformly distributed on $[0,1]$ 
There may be another approach: by symmetry, for given $\mu$ the expectation of the rounded value of $Y$ is $0$ and so the expectation of the rounded value of $X$ is $\mu$, meaning that the rounded value of $X$ (or the sample mean of several independent observations of it) is an unbiased estimator of $\mu$ 
