I'm given that $\sqrt{n}(\hat \theta_n - \theta) \rightarrow_d W \sim N(0,\theta^2)$ and that $\hat \theta_n \rightarrow_p \theta$. I am trying to find an approximation for $P(|\hat \theta_n -\theta|>k)$ that does not depend on $\theta$. Presumably it depends on $n$.

I know that $(\hat \theta_n - \theta) \rightarrow_d Z \sim N(0, \frac{1}{n}\theta^2)$. Is it justified to say that for a fixed $n$, $(\hat \theta_n - \theta)$ is distributed similarly to $N(0,\frac{1}{n}\theta^2)$?

My strategy was to employ Markov's or Chebyshev's Inequality, but I cannot get rid of $\theta$. I am questioning if there can be any meaningful approximation at all.

I believe $|\theta_n-\theta|$ will have folded normal distribution. It's mean will depend on variance, which depends on $\theta$, so Markov does not work.

Any help will be very appreciated! Thanks!

  • $\begingroup$ How about trying to use that $\sqrt{n} \frac{(\hat\theta_n-\theta)}{\theta}~N(0,1)$ and using the inequalities to approximate $\frac{\theta_n}{\theta}$ $\endgroup$ Nov 7, 2017 at 2:40
  • $\begingroup$ @DanielOrdoñez, I am not sure I understand. Could you try to explain furthwe? $\endgroup$
    – ElChorro
    Nov 8, 2017 at 0:23
  • $\begingroup$ Well if you have that $\sqrt{n}(\hat\theta_n-\theta)\sim N(0,\theta^2)$ then $\sqrt{n}(\frac{\hat\theta_n-\theta}{\theta})\sim N(0,1)$. And you can use this to conclude for example $P(|\sqrt{n}(\frac{\theta_n-\theta}{\theta})|\geq 1.96)=0.05$ $\endgroup$ Nov 8, 2017 at 0:43
  • $\begingroup$ I do know that, but as per question I want to find an approximation for $P(|\hat \theta_n -\theta|>k)$. First part of the question asked for an approximation that may depend on $\theta$ so using similar logic I concluded $P(|\hat \theta_n -\theta|>k) \approx 2\Phi(-\frac{\sqrt{n}}{\theta} 2) $ $\endgroup$
    – ElChorro
    Nov 8, 2017 at 0:48
  • $\begingroup$ Well from the above you can get: for example, $P(\hat\theta_n>k\theta)=0.025$ where $k=\frac{1.96}{\sqrt{n}}+1$, but I see your point finding a bound for the difference seems to be harder than this way... $\endgroup$ Nov 8, 2017 at 1:22


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