Bulletin Editor Anirban DasGupta writes:

The problem asked was to settle the possibility of consistent estimation with incomplete data in three examples, and to provide a concrete one when consistent estimation is possible.

Intuitively, in case (a), you can only infer about $|\mu |$, but not the sign of $\mu$. Denote $Y_i=|X_i|$; the distribution of each $Y_i$ and hence the joint distribution of $(Y_1,\dots ,Y_n)$ depends only on $|\mu |$. Suppose $c(Y_1,\dots ,Y_n)$ is a consistent estimate (tor) of $\mu$. Fix $\mu >0$. Then, $P_{\mu }(c(Y_1,\dots ,Y_n)<\mu /2)=P_{-\mu }(c(Y_1,\dots ,Y_n)<\mu /2)\to 1$, as $n\to \mathrm{\infty }$, which would contradict $c(Y_1,\dots ,Y_n)\stackrel{P}{⟹}\mu$ under every $\mu$.

In case (b), consistent estimation of $\lambda$ is possible. This is because the family of Poisson distributions is strictly MLR in $X$ and hence, for any $k$, ${\overline{F}}_k(\lambda )=P_{\lambda }(X>k)$ is strictly increasing in $\lambda$; it is also continuous. Denote $Y_i=I_{X_i>k}$. Then, $\overline{Y}$ is a consistent estimate of ${\overline{F}}_k(\lambda )$, and by the continuous mapping theorem, ${\overline{F}}_k^{-1}(\overline{Y})$ is a consistent estimate of $\lambda$; the estimate may be defined arbitrarily (for example, as 1), if $\overline{Y}=0$ or $n$.

Finally in case (c), perhaps a little surprisingly, consistent estimation of the exponential mean $\lambda$ is possible as well. By a direct calculation, the fractional part $Y=\{X\}$ has the density $g(y|\lambda )=\frac{1}{\lambda (1-e^{-1/\lambda })}{e}^{-y/\lambda },0<y<1$. This is a regular one parameter Exponential family and therefore the natural parameter $-\frac{1}{\lambda }$ is consistently estimable, and so, $\lambda$ is also consistently estimable. For example, a concrete consistent estimate can be constructed by estimating $E_{\lambda }(Y)$, necessarily a strictly increasing function of the natural parameter, by $\overline{Y}$, and then by using the inverse function.