Find min sufficient statistics for a continuous random sample

**tttcomrader** · September 9th 2011, 01:34 PM

Let $X_1,...,X_n$ be a random variable from an absolutely continuous distribution with density:

$f_ \theta (x) = \left \{ \begin {array}{rcl} \frac {2x}{ \theta ^2} & x \in (0, \theta ) \\ 0 & \mbox {otherwise} \end {array} \right.$

Find the minimum sufficient statistics T and its density.

The solution is as of follows:

Let $T(x) = \mbox {max} \{ X_i \} , M(x) = \mbox {min} \{ X_i \}$

Then $f_ \theta (x_1, \ldots ,x_n) = f _ \theta (x_1) \cdots f_ \theta (x_n)$

$= \frac {2x_1}{ \theta ^2} \cdots \frac {2x_n}{ \theta ^2} = \prod ^n_{i=1} \frac {2x_i}{ \theta ^2}$

$= \frac {2^n}{ \theta ^{2n}} \cdot \prod ^n_{i=1}x_i$

$= \frac {2^n}{ \theta ^{2n}} \cdot \prod ^n_{i=1} x_i \cdot 1 _{ \{ M(x)>0 \} } \cdot 1_{ \{ T(x)< \theta \} }$ , with $T = T(X)$

I understand that this density gives non zero values only when $M(X)>0$ and $T(X) < \theta$ but I don't understand how that last line with the two indicator functions come into being. Thank you!

**matheagle** · September 10th 2011, 08:56 PM

The suff stat is the max, because it cannot be separated from $\theta$
that's the factorization theorem in use.
NOW, for your answer.
The marginal densities are ZERO if the X's are not in $(0,\theta)$
HENCE, the joint density is ZERO if all the X's are not in $(0,\theta)$
And, note that

$0< X_1,...,X_n <\theta$ is the same as $0< X_{(1)}<\cdots <X_{(n)} <\theta$

is the same as $0< X_{(1)}$ and $X_{(n)} <\theta$

YOUR teacher should have had indicator functions on the previous lines as well, let me show you

**matheagle** · September 10th 2011, 09:06 PM

Your teacher left out the indicator functions on the first few steps
This should clear it up...

Originally Posted by tttcomrader

Let $X_1,...,X_n$ be a random variable from an absolutely continuous distribution with density:

$f_ \theta (x) = \left \{ \begin {array}{rcl} \frac {2x}{ \theta ^2} & x \in (0, \theta ) \\ 0 & \mbox {otherwise} \end {array} \right.$

Find the minimum sufficient statistics T and its density.

The solution is as of follows:

Let $T(x) = \mbox {max} \{ X_i \} , M(x) = \mbox {min} \{ X_i \}$

Then $f_ \theta (x_1, \ldots ,x_n) = f _ \theta (x_1) \cdots f_ \theta (x_n)$

$= \frac {2x_1}{ \theta ^2} I(0<x_1<\theta)\cdots \frac {2x_n}{ \theta ^2} I(0<x_n<\theta)$

$= \prod ^n_{i=1} \frac {2x_i}{ \theta ^2}I(0<x_i<\theta)$

$= \frac {2^n}{ \theta ^{2n}} \cdot \prod ^n_{i=1}x_i I(0<x_i<\theta)$

$= \frac {2^n}{ \theta ^{2n}} \cdot \prod ^n_{i=1} x_i \cdot 1 _{ \{ M(x)>0 \} } \cdot 1_{ \{ T(x)< \theta \} }$ , with $T = T(X)$

I understand that this density gives non zero values only when $M(X)>0$ and $T(X) < \theta$ but I don't understand how that last line with the two indicator functions come into being. Thank you!

Whoever did this doesn't completely understand.
The indicators were there from the beginning.
Sticking them at the end means they don't understand functions as well.

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