You observe a sample of 100 independent observations $X_i$ from a population with the density $$ g(x)=C \sqrt{\lambda} \exp \left(-\lambda x^2-\lambda^2 x^4\right), \quad-\infty<x<\infty $$ where $C=0.73077$ is the normalizing constant, and $\lambda>0$ is an unknown parameter.

(a) Find the minimal sufficient statistic for $\lambda$, and show that it is incomplete.

By the factorization theorem I was able to find that the minimal sufficient statistics are $T = (\sum_i^n x_i^2$, $\sum_i^n x_i^4)$.

I am not really sure how to prove that this is incomplete. My idea here is to make use of the second moment and fourth moment. The idea would be to find a function $u(T)$ such that $E(u(T)) = 0$ but $u(T) \ne 0$. But I am not too sure how to calculate the 2nd and 4th moments. Is there a common trick on how to calculate these integrals?

  • $\begingroup$ Consider changing the variable to $y = \lambda x^2.$ $\endgroup$
    – whuber
    Jul 20, 2023 at 16:18
  • $\begingroup$ When trying to find the second moment and considering the change of variable I end up with $\frac{C}{2\lambda} \int_{-\infty}^{\infty} \sqrt y ~exp(-y - y^2) dy$. I don't see how to move past this. $\endgroup$
    – Stats_Rock
    Jul 20, 2023 at 19:50
  • $\begingroup$ You won't succeed that way, because $y$ is never negative! If you really want an explicit expression for the moments, they can be written as linear combinations of Bessel $I_\alpha$ functions. But you might want to see whether you can carry out a demonstration without needing an explicit formula. $\endgroup$
    – whuber
    Jul 20, 2023 at 20:10
  • 1
    $\begingroup$ What I ended up doing is $E(X^2) = \frac{K^*}{2\lambda}$ for some constant K* and I also find that $E(X) = \frac{C^*}{2\sqrt{\lambda}}$ for some constant C*. And so $u(T) = S^2 - (2K^* - {C^*}^{2})(\frac{X^2}{2K^*})$ gives me what I want but I'm not sure if this is the correct approach. $\endgroup$
    – Stats_Rock
    Jul 20, 2023 at 20:34
  • $\begingroup$ @Stats_Rock $E(X)$ is obviously $0$ by symmetry, is there any typo? $\endgroup$
    – Zhanxiong
    Nov 6, 2023 at 15:30

1 Answer 1


First, let's calculate $E[X^2]$ and $E[X^4]$. By evaluating the integral with the given density directly, we have \begin{align} & E[X^2] = 2C\sqrt{\lambda}\int_0^\infty x^2\exp(-\lambda x^2- \lambda^2x^4)dx \\ =& 2C\sqrt{\lambda}\int_0^\infty \lambda^{-1}t\exp(-t-t^2)\frac{1}{2}\lambda^{-1/2}t^{-1/2}dt \tag{set $t = \lambda x^2$} \\ =& C_1\lambda^{-1}. \\ & E[X^4] = 2C\sqrt{\lambda}\int_0^\infty x^4\exp(-\lambda x^2 - \lambda^2x^4)dx \\ =& 2C\sqrt{\lambda}\int_0^\infty \lambda^{-2}t^2\exp(-t-t^2)\frac{1}{2}\lambda^{-1/2}t^{-1/2}dt \tag{set $t = \lambda x^2$} \\ =& C_2\lambda^{-2}. \\ \end{align} where $C_1 = C\int_0^\infty t^{1/2}\exp(-t - t^2)dt > 0$ and $C_2 = C\int_0^\infty t^{3/2}\exp(-t - t^2)dt > 0$ are constants that are independent of $\lambda$ (of course, you need to verify that these two integrals are indeed convergent, which is easy by comparing them with Gamma functions).

The expressions of $E[X^2]$ and $E[X^4]$, in particular, that the order of $E[X^4]$ happens to be a square of $E[X^2]$, inspire us to consider $g(x, y) = Ax^2 + By$, where $A, B$ are coefficients to be determined such that $E_\lambda\left[A\left(\sum X_i^2\right)^2 + B\sum X_i^4\right] = 0$ for all $\lambda > 0$. By independence, \begin{align*} & E\left[\left(\sum_{i = 1}^n X_i^2\right)^2\right] = nE[X^4] + n(n - 1)(E[X^2])^2 = (nC_2 + n(n - 1)C_1^2)\lambda^{-2}, \\ & E\left[\sum_{i = 1}^n X_i^4\right] = nE[X^4] = nC_2\lambda^{-2}. \end{align*} Therefore, by choosing $A = 1, B = -\frac{nC_2 + n(n - 1)C_1^2}{nC_2}$, we can have $E_\lambda\left[g\left(\sum X_i^2, \sum X_i^4\right)\right] = 0$ for all $\lambda > 0$ yet $g$ is clearly not zero. This proves that the family of given densities is not complete.

  • $\begingroup$ Thank you so much! This was very easy to follow. When I first attempted this I was able to find $E(X^2)$ and $E(X^4)$, but wasn't sure how to deal with the square term in the denominator. This was a brilliant way to make use of independence and the fact that $E(X^4)$ happens to be a square of $E(X^2)$. $\endgroup$
    – Stats_Rock
    Nov 8, 2023 at 2:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.