So say I have $x_1, x_2... x_n$ random sample with common pdf having a single unknown parameter $\theta$. The question is worded as follows: Find UMVUE for P(x>a). Is this problem just a matter of finding UMVUE for $\theta$ then substituting to the pdf?


The common pdf is $f(x)=\theta x^{-2}$ when $x>\theta$ and 0 otherwise.

I tried getting both the MLE and the Method of Moments estimator for $\theta$ but could not. I was able to get $P(x>a)$ as follows $\int_a^{\infty}\theta x^{-2}dx=\theta/a$ when $a>\theta$ and 1 otherwise. So, I need to find the UMVUE for $\theta/a$ and that is where I am stuck.


I have made further progress. Realizing that the minimum of the sample gives the only useful information about $\theta$. I obtained the distribution of the minimum $m$ as $f(m)=\frac{n\theta^n}{m^{n+1}}$. $E(m)=\frac{n}{n-1}\theta$ which means that $\frac{n-1}{n}m$ is unbiased for $\theta$.

My question is, can I say that the minimum is sufficient without using the factorization theorem and just arguing that given a sample, the most that can be drawn from it in relation to the parameter as defined is it's minimum? Also, is the minimum complete due to Pareto being exponential family?

  • $\begingroup$ Find a complete and sufficient statistics (CSS) for $\theta$. Then find an unbiased estimator of $\theta$ that is a function of the CSS. Using that, find an unbiased estimator of $\theta/a$ (which will also be a function of the CSS). Since this will be a function of the CSS, by Rao-Blackwell theorem, you get UMVUE. $\endgroup$ Aug 11, 2017 at 12:49
  • $\begingroup$ @Greenparker. Ok. I know that the joint pdf is $\theta^n\Pi_{i=1}^nx_i^{-2}$. This means that I can choose even a single $x_i$ as a sufficient statistic using the factorization theorem, right? $\endgroup$
    – user164144
    Aug 11, 2017 at 13:14
  • $\begingroup$ Ok.. so I just realized that the distribution of $x$ is $Pareto(\theta,1)$ which has an unbounded mean, which is why the typical strategies I tried failed. This is good news though since Pareto is a member of the exponential family which I think means that my sufficient stat $x_1$ is also complete. Anyone have a hint on how I can proceed with this? $\endgroup$
    – user164144
    Aug 11, 2017 at 13:42
  • $\begingroup$ Oh, just realized. Given a random sample, the minimum of the sample gives the only useful information about the parameter $\theta$ $\endgroup$
    – user164144
    Aug 11, 2017 at 13:49
  • $\begingroup$ That $\min X_i$ is complete for $\theta$ has to be shown separately. This Pareto pdf is not a member of exponential family as the support depends on $\theta$. $\endgroup$ Nov 16, 2019 at 20:17

1 Answer 1


Is this problem just a matter of finding UMVUE for $\theta$ then substituting to the pdf?

Not quite; because an unbiased estimate of $\theta$ may not yield an unbiased estimate of $P(X > a)$; unbiasedness is essential for a statistic to be UMVUE.

Instead, if $f(x)$ is the pdf of $X$, then $P(X > a)$ is $$P(X > a) = \int_a^{\infty} f(x) dx $$

$f(x)$ is a function of $x$ that also depends on $\theta$. When $x$ is integrated out, what will remain is a quantity that is dependent on the known $a$ and the parameter $\theta$. Thus,

$$P(X > a) = \int_a^{\infty} f(x) dx = g_a(\theta)\,.$$

Thus, you have to find the UMVUE of $g_a(\theta)$ (Which for example could look something like $(\theta - a)^2$ or $\sqrt{\theta(a - \theta)^2}$ etc.)

A way to tackle the problem would be to first find an unbiased estimator of $g_a(\theta) = P(X > a)$. Hint: $P(X >a) = E[I(X > a)]$.

  • $\begingroup$ Ok. So I was able to compute P(X>a) and got $\frac{\theta}{a}$. I am not sure how to use the hint. I am a bit confused on finding the UMVUE for an expression instead of just a single parameter. $\endgroup$
    – user164144
    Aug 11, 2017 at 11:04
  • $\begingroup$ @user164144 Assuming $\theta/a$ is correct, now you need to find the UMVUE for $\theta/a$. The first thing you need is an unbiased estimator of $\theta/a$, and then you need to show that this estimator has the uniformly minimum variance. Often, it is helpful to use an estimator that is a function of a sufficient and complete statistics. $\endgroup$ Aug 11, 2017 at 11:25
  • $\begingroup$ I get that as I have been doing that same procedure for other problems. However, this is the first time I have come across a problem where what is to be estimated is more complicated. Hmm.. should I treat this as a parameter and somehow make it appear in the original pdf of x? $\endgroup$
    – user164144
    Aug 11, 2017 at 11:59
  • 1
    $\begingroup$ Here are some other problems on this website that you may find helpful: This, this, this. $\endgroup$ Aug 11, 2017 at 12:05
  • $\begingroup$ I looked at the examples but don't see a connection to my current problem. Basically my problem is that if what is just being asked for is to find a UMVUE for $\theta$, I can do that using the procedure you described. However, what stumps me is that I have to find the UMVUE for $P(X>a)$ which by integrating the pdf from $(a,\infty)$ I know to be $\theta/a$ $\endgroup$
    – user164144
    Aug 11, 2017 at 12:12

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