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Suppose $\hat{\theta}_1=\int_0^x \frac {dF_n(y)} {1-F_n(y)}$ and $\hat{\theta}_2=\int_0^x \frac {f_n(y)} {1-F_n(y)} dy$ are two estimators of cumulative hazard function, where $f_n(y) =\frac {1} {b}\int k\left(\frac {y-u} {b}\right) dF_n(u)$ is an estimate of the density $f(y)$ by assuming $b\rightarrow 0$ with $bn\rightarrow \infty$ as $n\rightarrow \infty$, and $F_n(y)=\frac {1} {n}\sum_{i=1}^n I\left\lbrace X_i\leq y \right\rbrace$ is the empirical distribution function.

Is it possible to show that $\hat{\theta}_1$ converges to $\hat{\theta}_2$ in probability?

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  • $\begingroup$ That result is not possible unless you allow $b\to 0$ as $n$ increases. Otherwise, $f_n$ converges to a smoothed version of $f$ rather than the correct version of $f$. Assuming, then, that $b$ does shrink to zero, exactly how do you determine $b$ in terms of $n$? $\endgroup$ – whuber Oct 13 '16 at 18:28
  • $\begingroup$ Yes, you are right! I forgot to mention. $\endgroup$ – user00110014 Oct 13 '16 at 18:49

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