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Let $X_1,\ldots,X_n$ be independent random variables with densities:

$$f_{X_k}(x\mid\theta)= \begin{cases} e^{k\theta-x}& x\geq k\theta\\ 0& \text{otherwise} \end{cases} $$

Find the pmf for $T=\min_{k}\left(\frac{X_k}{k}\right)$.

So given the pdf of $X_k$, the pdf of $Y_k=\frac{X_k}{k}$ is given by

$$f_{Y_k}(y\mid\theta)=\begin{cases} ke^{k(\theta-x)}& x\geq \theta\\ 0& \text{otherwise} \end{cases} $$

Thus for each $k$, $1\leq k\leq n$, I need to find $\text{P}(T=Y_k).$ This is the same as calculating the probability that the remaining random variables are all greater than $Y_k$ for some fixed $Y_k=y_k$, and then integrating over all possible values of $Y_k$ weighted according to its pdf, where by independence, $$\text{P}(Y_1,\dots,Y_{k-1},Y_{k+1},\dots,Y_n > y_k)=\prod_{j=1,\dots,k-1,k+1,\dots,n}\int_{y_k}^{\infty}je^{j(\theta-y)} \, dy=\prod_{j=1,\dots,k-1,k+1,\dots,n}e^{j(\theta-y_k)}.$$

Thus we have:

$$\text{P}(T=Y_k)=\int_{\theta}^{\infty}\Big(\prod_{j=1,\dots,k-1,k+1,\dots,n}e^{j(\theta-y_k)}\Big)ke^{k(\theta-y_k)} \, dy_k = k\int_{\theta}^\infty e^{\frac{n(n+1)}{2}(\theta-y_k)} \, dy_k=\frac{2k}{n(n+1)}.$$

Whew! Ok this sums to $1$ as $k$ ranges between $1$ and $n$, which gives me confidence I did this correctly. But on the other hand, it's now requested that I show that $T$ is a sufficient statistic for the parameter $\theta$. Which by a theorem in the book means that the pdf of the joint distribution $(X_1,...,X_n)$, divided by the pmf of $T$, should be a function which is independent of $\theta$.

But the pmf of $T$ is itself independent of $\theta$, that would mean that the pdf of $(X_1,\dots,X_n)$ would also have to be independent of $\theta$, which doesn't appear to be true, since by independence it's just the product of the individual density functions and $\theta$ does not disappear upon taking that product.

The book uses a different and much easier theorem to prove that $T$ is a sufficient statistic, so I'm not looking for a proof of that, but instead would like to find out what's wrong with my method. Thanks.

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You have not yet found the correct expression . You are asked to find $P(T=x)$ for any number $x$, whereas you currently have $P(T=Y_k)$, that is the probability that the minimum is achieved at a particular $Y_k$. –  Alex R. Aug 14 '13 at 23:15
    
@AlexR. so you're saying that $T$ is not a random variable taking values in $1,...,n$ depending on which $X_k$ is the smallest, but instead takes the value of $X_{(1)}$? –  cactuar Aug 14 '13 at 23:18
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Yes I believe so. Usually for your interpretation, instead of seeing $\min_k$ you'd see $\mbox{argmin}_k$ which returns the index of the minimizer. –  Alex R. Aug 14 '13 at 23:23
    
@AlexR. ahh ok, well that would explain my inconsistent results, thanks. –  cactuar Aug 14 '13 at 23:26
    
I mangled my first answer by overlooking something; I've deleted it and posted a second one. Two small points: It's a pdf rather than a pmf; the distribution is not discrete. (The distribution of the particular index $k$ for which the minimum is realized is of course discrete.) Also, you've got $f_{Y_k}(y)=$ a function of $x$, not of $y$. The same variable should appear on both sides. I will extend my second answer somewhat, since it doesn't answer everything that was asked. –  Michael Hardy Aug 15 '13 at 1:41

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up vote 1 down vote accepted

Sorry, I missed something and got my last answer mangled. I've deleted that.

Notice that $$ \Pr(Y_k>x) = \int_x^\infty ke^{k(\theta-w)} \,dw = e^{k(\theta-x)}. $$ So \begin{align} \Pr(\min>x) & = \Pr(\text{for all $k$, }Y_k>x) \\[10pt] & = e^{\theta-x} e^{2(\theta-x)} e^{3(\theta-x)}\cdots e^{n(\theta-x)} \\[10pt] & = e^{n(n+1)(\theta-x)/2}. \end{align} Consequently for $x>\theta$, we have $$ f_\min(x) = \dfrac{d}{dx}(1- e^{n(n+1)(\theta-x)/2}) = \frac{n(n+1)}2\cdot e^{n(n+1)(\theta-x)/2}. $$ Dividing by $e^{\theta-x} e^{2\theta-x} e^{3\theta-x}\cdots e^{n\theta-x}$ $= e^{n(n+1)\theta/2 - nx}$, one sees that the $\theta$ cancels and the $x$ doesn't.

Here's what seems to be your error: You wrote:

But the pmf of $T$ is itself independent of $\theta$

That's not true. The pdf is derived above. What you found was not the pdf of $T$, but the pmf of the discrete random variable whose value is the index $k$ for which $T=Y_k$. The random variable $T$ itself is a continuous random variable. Maybe you're confusing $T$ with that discrete random variable.

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