1
$\begingroup$

Let $X_1, ..., X_n$ be a sample from the continuous density $C~exp(-(x-\theta)^4)$ (for $ -\infty < x < \infty$) with $\theta$ as unknown parameter. Show that the minimal sufficient statistic is incomplete.

By factorization theorem we see that $T = [\sum_i^n x_i, \sum_i^n x_i^2, \sum_i^n x_i^3]$.

We also can find that $E(X) = \theta$, $~E(X^2) = K + \theta^2$, and $E(X^3) = 3K\theta + \theta^3$, for some constant K.

Also note that: $Var(X) = E(X^2) - E(X)^2 = K = E(S^2)$ and $E(\bar{X}) = \theta$.

Now to show completeness, I want to find an arbitrary function $u(T)$, such that $E(u(T)) = 0$ but $P(u(T) = 0) \ne 1$

Now when choosing u(T), do I need to incorporate all three sufficient statistics? or could I use just two out of the three? Here is my attempt with all three:

is $u(T) = S^2 - K + \frac{\sum_i^n X_i^3}{n} - 3K \bar{X} - (X_1)(X_2)(X_3)$ a valid function to use

since $E(u(T)) = E(S^2) - K + \frac{nE(X^3)}{n} - 3K E(\bar{X}) - E(X_1)E(X_2)E(X_3) = 0$ ?

Note that: ($S^2$ is a function of $T_1$ and $T_2$, and $\frac{\sum_i^n X_i^3}{n}$ is a function of $T_3$)

Is there another way to show it is incomplete?

$\endgroup$
4
  • 2
    $\begingroup$ A function $u(T)$ turns its entry $T$ into a real number. This does not mean it must involve all three components of $T$. $\endgroup$
    – Xi'an
    Commented Jun 18, 2023 at 8:05
  • 1
    $\begingroup$ For instance, $U(T)=S^2-K$... $\endgroup$
    – Xi'an
    Commented Jun 18, 2023 at 16:44
  • $\begingroup$ Thank you, this was very helpful! Another slightly off-topic question I had is: I am able to show that this is a regular class of the exponential family, doesn't this suggest that the sufficient statistic is Complete and Minimal? But we have just shown that it is incomplete. $\endgroup$
    – Stats_Rock
    Commented Jun 18, 2023 at 18:54
  • 2
    $\begingroup$ Please abstain from asking new questions thru comments, better ask a new question! $\endgroup$
    – Xi'an
    Commented Jun 18, 2023 at 20:53

1 Answer 1

2
$\begingroup$

There's no need to use all 3 components of $T$. One function that works is $u(T)=nT_2-T_1^2-K.$

$\endgroup$