# Showing incompleteness of density

You observe a sample of 100 independent observations $$X_i$$ from a population with the density $$g(x)=C \sqrt{\lambda} \exp \left(-\lambda x^2-\lambda^2 x^4\right), \quad-\infty where $$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?

• Consider changing the variable to $y = \lambda x^2.$
– whuber
Jul 20, 2023 at 16:18
• 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. Jul 20, 2023 at 19:50
• 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.
– whuber
Jul 20, 2023 at 20:10
• 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. Jul 20, 2023 at 20:34
• @Stats_Rock $E(X)$ is obviously $0$ by symmetry, is there any typo? Nov 6, 2023 at 15:30

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.
• 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 $E(X^2)$. Nov 8, 2023 at 2:40