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Let $X_1, \cdots, X_n$ be i.i.d. $Unif(0,\theta)$ and $T = \max\{X_1,X_2,···,X_n\}$. Show that T is a sufficient statistic using the definition. So I need to show that for $t>0$, $\Bbb P(X_1 \leq x_1, \cdots, X_n \leq x_n \lvert T \leq t)$ does not depend on $\theta$. Here are my computations :

$$\Bbb P(X_1 \leq x_1, \cdots, X_n \leq x_n \lvert T \leq t)=\frac{\Bbb P(X_1 \leq x_1, \cdots, X_n \leq x_n , T \leq t)}{\Bbb P( T \leq t)}$$

$$=\frac{\Bbb P(X_1 \leq x_1, \cdots, X_n \leq x_n , T \leq t)}{\Bbb P( T \leq t)}=\frac{ \Bbb P(X_1 \leq \min(x_1,t), \cdots, X_n \leq \min(x_n,t))}{\Bbb P(X_1 \leq t, \cdots, X_n \leq t)}$$

$$=\frac{ \Bbb P(X_1 \leq \min(x_1,t), \cdots, X_n \leq \min(x_n,t))}{\Pi_{i=1}^n \Bbb P(X_i \leq t)}=\frac{\Pi_{i=1}^n \int_{-\infty}^{\min(x_i,t)}\Bbb 1_{[0,\theta]}dx}{\bigg(\int_{-\infty}^{t}\Bbb 1_{[0,\theta]}dx\bigg)^n}$$

It is very technical but we can finally write it as $$\Bbb P(X_1 \leq x_1, \cdots, X_n \leq x_n \lvert T \leq t)=\frac{\Pi_{i=1}^n \min(x_i,t,\theta)}{(\min(t,\theta))^n}$$

Let us suppose for example that $t\geq \theta$ and all $x_i$ are also bigger than $\theta$ then $\Bbb P(X_1 \leq x_1, \cdots, X_n \leq x_n \lvert T \leq t)=\frac{\theta^n}{\theta^n}=1$ which does not depend on $\theta$.

The big issue is starting to be clear : if $t \geq \theta$ and (at least) one of $x_i$ is smaller than $\theta$ then we have that $\Bbb P(X_1 \leq x_1, \cdots, X_n \leq x_n \lvert T \leq t)=\frac{x_i \theta^{n-1}}{\theta^n}=\frac{x_i}{\theta}$ that depends on $\theta$ !

If maths are never wrong, then I am but where ? I was told there must be a mistake in my computations but they seem ok. Can anyone see what's not right ?

(I know I can use the equivalent factorisation but I really want to do it by definition).

Kilkik
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    In fact with probability $1$, all the $X_1$ are less than or equal to $\theta$. So too is $T$ – Henry Mar 15 '22 at 15:05
  • HINT: can you simplify $\min\left{x_i,t\right}$? – DeBARtha Mar 15 '22 at 15:08
  • I think you want to show $\Bbb P(X_1 \leq x_1, \cdots, X_n \leq x_n \mid T = t)$ does not depend on $\theta$, not $\cdots \mid T \le t)$ – Henry Mar 15 '22 at 15:17
  • @DeBARtha I can give the formal expression with absolute values but it is not simplified a lot lol – Kilkik Mar 15 '22 at 15:19
  • @Henry That was in fact my first guess but it seems to lead to same calculations but a $X_i$ will be equal to $t$ no ? – Kilkik Mar 15 '22 at 15:20
  • The conditional distribution is not pretty: if $n=3$ then it is constrained to the three outer faces of a cube $[0,t]^3$. But it does not depend on $\theta$ – Henry Mar 15 '22 at 15:29

1 Answers1

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Note that with prob. 1 we have $X_i\leq T\leq \theta$, and thus you only need consider the case $x_i\leq t\leq \theta $ under which your expression simplifies to $\prod_i x_i\over t^n$, which is independent of $\theta$ as desired.

You can also directly appeal to this definition of sufficiency applied for absolutely continuous random variables:

$${L(x_1,...,x_n;\theta)\over f_T(t;\theta)},\quad t=T(x_1,...,x_n)$$

is independent of $\theta$, where $L(x_1,...,x_n;\theta)$ is the likelihood and $f_T$ is the density of statistic $T$.

In your setup, we have

$$\small{L(x_1,...,x_n;\theta)\over f_T(t;\theta)}={{1\over \theta^{n}}\prod_i {\bf 1}_{0\leq x_i\leq \theta}\over {nt^{n-1}\over \theta^n}{\bf1}_{0\leq t\leq \theta}}={{1\over \theta^{n}}{\bf 1}_{\max\{ x_1,...,x_n\}\leq \theta}\prod_i {\bf 1}_{0\leq x_i} \over {nt^{n-1}\over \theta^n}{\bf1}_{0\leq t\leq \theta}}={{1\over \theta^{n}}{\bf 1}_{t\leq \theta}\prod_i {\bf 1}_{0\leq x_i} \over {nt^{n-1}\over \theta^n}{\bf1}_{0\leq t }{\bf1}_{t \leq \theta}}={\prod_i {\bf 1}_{0\leq x_i} \over {nt^{n-1}}{\bf1}_{0\leq t }},$$

which is independent of $\theta$ as desired.

Golden_Ratio
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