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 $(Y_1, \ldots , Y_n)$ depends only on $|\mu |$. Suppose $c(Y_1,\ldots , Y_n)$ is a consistent estimate (tor) of $\mu$. Fix $\mu > 0$. Then, $P_{\mu }(c(Y_1,\ldots , Y_n) < \mu /2) = P_{-\mu }(c(Y_1,\ldots , Y_n) < \mu /2) \to 1$, as $n \to \infty$, which would contradict $c(Y_1,\ldots , Y_n) \stackrel {P} {\Longrightarrow } \mu$ 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 $k$, $\bar{F}_k(\lambda ) = P_\lambda (X > k)$ is strictly increasing in $\lambda$; it is also continuous. Denote $Y_i = I_{X_i > k}$. Then, $\bar{Y}$ is a consistent estimate of $\bar{F}_k(\lambda )$, and by the continuous mapping theorem, $\bar{F}_k^{-1}(\bar{Y})$ is a consistent estimate of $\lambda$; the estimate may be defined arbitrarily (for example, as 1), if $\bar{Y} = 0$ 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 $\bar{Y}$, and then by using the inverse function.