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Someone had asked this earlier, but since it was good practice for my qualifying exam coming up, I figured I would ask and share my work on the problem. The problem is:

Suppose $X$ is Unif$(0,\theta)$. Let $T = X_{(1)}$ (minimum order statistic). Is $T$ complete?

My initial thought was no since usually lectures show $X_{(n)}$ is, but here was my attempt:

I assume that $E[g(T)]=0$ for some function $g$ and want to find if $P(g =0)=1$.

$$P(T \leq t) = 1 - P(T > t) = 1-\left(1-\frac{t}{n} \right)^n$$

So taking derivatives to get the pdf:

$$f_T(t) = \left(1-\frac{t}{n}\right)^{n-1}$$

So now working with expectations:

$$0=E[g(T)] = \int_0^\theta g(t)\left(1-\frac{t}{n}\right)^{n-1} dt$$

This integral is differentiable, and since the integral is $0$, then the derivative is $0$, so taking the derivative with respect to $\theta$:

$$0= g(\theta)\left(1-\frac{\theta}{n}\right)^{n-1}$$

Hence $g(\theta) =0$, so $T$ is a complete statistics.

Does this argument work? I basically modeled it after the proof of showing the max order statistic is complete. If $X_{(1)}$ is not complete, where does the proof go wrong?

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  • $\begingroup$ First mistake: one wants to prove that $P_\theta(g(T) =0)=1$ for every $\theta$ ($g(T)$, not $g$). Second mistake: $P_\theta(T \leqslant t) = 1-\left(1-\frac{t}{\theta} \right)^n$ for every $t$ in $(0,\theta)$ ($\frac{t}{\theta}$, not $\frac{t}n$). Can you correct these and pursue? $\endgroup$ – Did May 21 '16 at 19:19
  • $\begingroup$ @Did Thanks for the corrections! So using the fixed pdf and getting to the integral, I get $0=E[g(T)] = \int_0^{\theta}g(t)\frac{n}{\theta}\left(1-\frac{t}{\theta} \right) dt$. So now when I take a derivative with respect to $\theta$, the $\left(1-\frac{t}{\theta} \right)$ term becomes $0$, so $g(T)$ does not have to be $0$ with probability $1$, and thus it's not a complete statistic? $\endgroup$ – Brenton May 21 '16 at 19:28
  • $\begingroup$ Huh? First, this is $\left(1-\frac{t}\theta\right)^{n-1}$, not $\left(1-\frac{t}\theta\right)$. Second, the derivative of the integral is not what you say. $\endgroup$ – Did May 21 '16 at 19:40
  • $\begingroup$ @Did Whoops, left out the $n-1$. So I have $\int_0^{\theta} g(t)\frac{n}{\theta}\left(1-\frac{t}{\theta}\right)^{n-1} dt$. So taking the derivative w.r.t $\theta$... don't I get $g(\theta)\frac{n}{\theta}\left(1-\frac{\theta}{\theta} \right)^{n-1}$? $\endgroup$ – Brenton May 21 '16 at 19:45
  • $\begingroup$ No. What is the formula for $$\frac{d}{d\theta}\left(\int_{a(\theta)}^{b(\theta)}h(\theta,t)dt\right),$$ already? $\endgroup$ – Did May 21 '16 at 19:47

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