Consistent estimator and distribution function

a general question: If the distribution function $$F_n$$ of some estimator $$T_n$$ suffices \lim_{n \rightarrow \infty} F_n(x) = 1 \text{ or } 0 \forall x}.

Does that imply that $$T_n$$ is consistent?

I am not sure if this question in general even makes sense. But in this paper https://pdfs.semanticscholar.org/3d64/5e60691838bf4699e79458d96930ba7bf24e.pdf In the proof of theorem 1 the author uses this argument if I understand it correctly.

Updated question:

Let $$n=b^k$$ with $$b \geq 3$$ and odd. And let $$x_1, \dots x_n$$ be independent observations. The sample median of $$b$$ observations is denoted by $$T_b$$. The remedian $$T_{b^k} = T_b(T_{b^{k-1}}, \dots, T_{b^{k-1}})$$. The author claims: If $$F$$ has a continuous density $$f$$ that is strictly positive at $$\text{Med(F)}$$. Then the remedian is consistent for $$\text{Med(F)}$$.

The proof of the author: Let $$b=2m+1$$. The distribution function of the sample median is: $$G_b(x) = R_b(F(x))$$ where $$R_b(u) = \sum_{m+1}^b \binom{b}{j}u^j(1-u)^{b-j}$$ Therefore the distribution function the the remedian is: $$G_{b^k} = R_b(R_b(\dots R_b(F(x)))) = R_b^{(k)}(F(x))$$ He shows that $$R_b$$ is strictly convex for $$u \leq \frac{1}{2}$$ and strictly concave for $$u \geq \frac{1}{2}$$.

So far I understand it.

But then he shows that:

$$\lim_{k \rightarrow \infty} G_{b^k}(x) = 1$$ for $$x$$ such that $$F(x) > \frac{1}{2}$$ and

$$\lim_{k \rightarrow \infty} G_{b^k}(x) = 0$$ for $$x$$ such that $$F(x) < \frac{1}{2}$$

and then concludes that this proofs the consistency of $$T_{b^k}$$

Which I don't understand. And where is the strictly positive $$f$$ at $$\text{Med(F)}$$ needed?

• Presumably by "suffices" you mean "satisfies." Have you tried applying a definition of consistency? What definition are you using? It may be of interest to note there are consistent estimators that do not have this limiting property. For instance, the sample mean of any continuous finite-variance distribution with mean $\mu$ does not have this property (because the limiting value of $F_n(\mu)$ is $1/2$). – whuber May 12 at 17:03
• @whuber sorry my initial question was wrong, I updated it so it makes more sense I hope – user1970122 May 12 at 17:34
• The expression $(5.2)$ for the distribution function of the median is correct only for continuous distributions and is asymptotically correct (in probability) only for distributions continuous in a neighborhood of their median. – whuber May 12 at 19:15
• I think I got it, I just have to use the definition of consistency: $P(|T_{b^k}-\text{med(F)}| \geq \epsilon) = 1-G_{b^k}(\epsilon + \text{med(F)}) = 0$ as $k \rightarrow \infty$ and if $T_{b^k} \geq \text{med(F)}$ because of the first limit stated in the question. The other case works similar because of the second limit. I still dont quite understand why $f$ needs to be striclty positive at $\text{med(F)}$ – user1970122 May 13 at 10:25
• Your formula reflects the right understanding, but it needs modification to $$P(|T_{b^k}-\operatorname{med}(F)|\gt\epsilon)\le 1-G_{b^k}(\epsilon+\operatorname{med}(F))+G_{b^k}(-\epsilon+\operatorname{med}(F)){\to}_{k\to\infty}(1-1)+0=0.$$ To see why continuity at the median is needed, consider the simplest possible $F$ that is not continuous at the median, such as a Bernoulli$(p)$ distribution with $p\ne1/2.$ The sample median can only equal $0$ or $1,$ whence it must have a discrete distribution and therefore it cannot be given by $R_b.$ – whuber May 13 at 15:16