4

Lex $X_1, X_2 \sim U(0, \theta)$ i.i.d , then obviously the sufficient statistic would be $T = \max\{X_1, X_2\}$, which can be easily proven by factorization theorem. However, I try to compute the density $f(x_1, x_2\mid t)$, which turns out to depend on the parameter $\theta$. This absolutely contradicts the definition of sufficient statistic:

$$ f(x_1, x_2, t) = \frac 1 {\theta^2}I\{t\leq \theta, \max\{x_1, x_2\}=t\} $$

$$ f(t) = \frac{2t}{\theta^2}I\{t \leq \theta\} $$

Thus

$$ f(x_1, x_2\mid t) = \frac 1 {2t} I\{t\leq \theta\} $$

which depends on $\theta$ through the indicator function $I\{t\leq \theta\}$.

What's wrong with my reasoning? Thanks a lot!

Michael Hardy
  • 1
Xiaojie
  • 43
  • 2

2 Answers2

1

I think you should have $$f(x_1,x_2 \mid t) = \frac{1}{2t} I\{0 \le t \le \theta, \max\{x_1,x_2\} = t\} = \frac{1}{2t} I\{\max\{x_1,x_2\} = t\}.$$

Regarding the indicator arithmetic: if $I\{0\le t \le \theta\} = 0$, then $f(x_1,x_2 \mid t)$ is not even a well-defined probability distribution, so I think you can ignore this case and assume $0 \le t \le \theta$ so that the denominator is nonzero.

  • if $I\{0 \le t \le \theta, \max\{x_1,x_2\}=t\}=1$, then $I\{0 \le t \le \theta\} =1$, so the ratio is $1$.
  • otherwise $I\{0 \le t \le \theta, \max\{x_1,x_2\}=t\}=0$, so the ratio is $0$ regardless of what the denominator is (even when the denominator is zero)

Edit:

$$P(T \le t) = (t/\theta)^2 I\{0 \le t \le \theta\} \implies f(t) = \frac{2t}{\theta^2} I\{0 \le t \le \theta\}$$

$$f(x_1,x_2,t) = \frac{1}{\theta^2} I\{0 \le t \le \theta, \max\{x_1,x_2\}=t\}$$

angryavian
  • 89,882
  • But it should be $1/t^2$, not $1/(2t). \qquad$ – Michael Hardy Jan 07 '17 at 21:54
  • @MichaelHardy Why is that? Isn't $f(t)$ the density of $T$, not the cdf? – angryavian Jan 07 '17 at 22:46
  • Thanks for your answer. I think maybe it's because I don't understand very well the definition of conditional density $f(x_1, x_2|t)$ at points where the denominator is 0. In your answer you seems to cancel out the $I{0 \leq t \leq \theta}$ occurring in both the numerator and denominator. I can sort of accept this. But how about the general case like when the denominator and numerator are both equal zero but they are different in form? How should we deal with it then? @angryavian – Xiaojie Jan 08 '17 at 15:04
  • @Xiaojie see my edit – angryavian Jan 08 '17 at 16:14
  • @angryavian I think your edit may not justify the result very well. According to your second point, if that the numerator is 0 leads to ratio 0, then $t>\theta$ means numerator is 0, which implies the ratio thus the result is 0. This means the result will depend on $\theta$. Perhaps it's still because the numerator and the denominator depends on $\theta$ in the same manner so we can sort of cancel them. – Xiaojie Jan 08 '17 at 20:40
  • @Xiaojie I edited again. I think for the conditional distribution, you should ignore the case $t > \theta$. (But it is important for the joint distribution and marginal distribution.) – angryavian Jan 08 '17 at 20:57
1

$$f(x_1,x_2\mid t)\overset{\text{Bayes}}{=}\frac{f(x_1,x_2,t)}{f(t)}=\frac{f(x_1,x_2)\mathbb{I}[x_1,x_2\le t]}{\mathbb{P}[x_1,x_2\le t]}=\frac{\frac{1}{\theta^2}\mathbb{I}[x_1,x_2\le t]}{\left(\frac t \theta \right)^2} = \frac{\mathbb{I}[x_1,x_2\le t]}{t^2}$$

Michael Hardy
  • 1
πr8
  • 10,800
  • $+1,$ but with an objection: The widepspread practice of using the same symbol, $f$, to refer to various different functions is wrong. There is a difference between $f_X(3)$ and $f_Y(3),$ the values of different functions at the same particular value of the argument. One should write $f_{X_1,X_1,\mid, T=t}(x_1,x_2) = \cdots$ etc. – Michael Hardy Jan 07 '17 at 22:51
  • @MichaelHardy thanks, and I tend to agree ... I lean towards using more concrete expressions ($P(X<x)$, for example, as opposed to densities) where possible. I'm happy to amend with the subscripts if you feel that would be best for understanding. – πr8 Jan 07 '17 at 22:55