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I am stuck on a problem for my Statistical theory class. So the problem goes like this:

Let X be the discrete uniform random variable, namely, X has the pmf:

$f(x)=\frac{1}{\theta}, x=1,2,...,\theta$

and we have a sample of size n, ${X_1,...,X_n}$.

Just like in the continous case, $T=Max(X_1,...,X_n)$ a) Show that T is a complete sufficient statistic for $\theta$.

So I have to figure out what E(g(t)) is since we are going about finding a complete sufficient statistic using the definition.

What I am stuck on is how do I figure this out since this is a discrete uniform distribution and not a continuous one.

What I have so far is this:

$E(g(T))=\Sigma g(t)*f_t(t)$

Which I am not sure if this is right since I know that $f_t(Max)=\frac{nt^{n-1}}{\theta^n}$.

Would I just figure out $E(g(T))=\Sigma g(t)*\frac{nt^{n-1}}{\theta^n}$?

So then I would get:

$E(g(T))=\frac{n}{\theta^n}\Sigma g(t)*t^{n-1}$

Let $h(t) =\frac{g(t)}{t}$

$=\frac{n}{\theta^n}\Sigma h(t)t^n$

The problem that I know this blows up to infinity and does not go to 0 like I need it to go.

Could someone please check my work and let me know if I am going wrong somewhere with this problem?

Thanks!

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  • $\begingroup$ It seems like you are mixing up the probability mass function of the maximum of $n$ independent discrete random variables uniformly distributed on the finite set $\{1,2,\ldots, \theta\}$ with the probability density function of the maximum of $n$ independent continuous random variables uniformly distributed on the finite interval $[0,\theta]$ $\endgroup$ – Dilip Sarwate Apr 11 '13 at 3:01
  • $\begingroup$ So what would be the pmf for the max on a finite interval? The whole discrete uniform distribution thing has been throwing me off. $\endgroup$ – Perdue Apr 11 '13 at 14:47
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    $\begingroup$ If $1 < a \leq \theta$ and exactly one or two or three or ... or $n$ of the $X_i$'s equal $a$ while the remaining $X_j$ are smaller than $a$, then it must be that $\max_i X_i = a$, right? So, maybe $$P\{\max_i X_i = a\} = \sum_{k=1}^n \binom{n}{k}\left(\frac{1}{\theta}\right)^k\left(\frac{a-1}{\theta}\right)^{n-k},~~ 1 < a \leq \theta ?$$ I will leave it to you to work out $P\{\max_i X_i = 1\}$. $\endgroup$ – Dilip Sarwate Apr 11 '13 at 14:58
  • $\begingroup$ Isn't this pretty much the binomial distribution? $\endgroup$ – Perdue Apr 11 '13 at 15:46
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    $\begingroup$ >Isn't this pretty much the binomial distribution? Well, to parody a quote from an ex-President, it pretty much depends on what the meaning of pretty much is. $\endgroup$ – Dilip Sarwate Apr 11 '13 at 16:46

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