# Minimal sufficiency and UMVUE in a pseudo-Normal distribution

I already asked a (stupid) question about this problem here thinking I wouldn't have problems to continue it but I was pretty wrong. I'm finding several more problems trying to solve it. I'll try to summarize what I've done so you don't get bored. The problem is:

Let X be a random absolutely continuous variable with probability density function $$f_{\lambda\mu}(x) = \sqrt{\frac{\lambda}{2\pi x^3}}\exp{\left\{-\frac{\lambda}{2\mu^2x}(x-\mu)^2\right\}} \quad x>0$$ with $\mu,\lambda>0$. Find the MLE ($T_1$) of $\mu$ and ($T_2$) of $1/\lambda$ for a sample of size $n$. Study their minimal sufficiency. Provided T_2 is complete and $\lambda n T_2\to \chi^2_{n-1}$ find the UMVUE of $1/\lambda$.

Likelihood function: $$f_{\lambda\mu}(x_1,\ldots,x_n) = \left( \frac{\lambda}{2\pi}\right)^{n/2}\prod x_i^{-3/2}\exp{\left\{ -\frac{\lambda}{2\mu^2}\sum\frac{(x_i-\mu)^2}{x_i} \right\}}\quad x_1,\ldots,x_n >0$$

I've started finding the MLE of $\mu$ and $1/\lambda$, which are:

$$T_1(x_1,\ldots,x_n) = \hat\mu = \frac{1}{n} \sum_{i=1}^n x_i = \bar x$$ and $$T_2(x_1,\ldots,x_n) = \frac{1}{\hat \lambda} = \frac{1}{n\hat\mu^2} \sum_{i=1}^n \frac{(x_i-\hat\mu)^2}{x_i}$$ According to Fisher-Neyman's factorization theorem, $\hat\mu$ is not sufficient because I cannot factorize the likelihood function acordingly, thus it is not minimal sufficient.

# The first part:

I've tried to check that $\hat\lambda$ is minimal sufficient saying that the expression

$$\frac{f_\lambda (x_1,\ldots,x_n)}{f_\lambda (x_1',\ldots,x_n')} = \frac{\prod x_i^{-3/2}}{\prod x_i'^{-3/2}}\exp\left\{ \frac{\lambda}{2\mu^2}\left[ \sum\frac{(x_i'-\mu)^2}{x_i'} - \sum \frac{(x_i-\mu)^2}{x_i}\right] \right\}$$ will not be in terms of $\lambda$ if, and only if, $$\sum\frac{(x_i'-\mu)^2}{x_i'} = \sum \frac{(x_i-\mu)^2}{x_i}$$ So $\sum \frac{(x_i-\mu)^2}{x_i}$ is a minimal sufficient estimator of $1/\lambda$, but $T_2$ is not. Is this correct?

# The second part:

After this I'm supossed to find the UMVUE of $1/\lambda$, supposing $T_2$ as complete and knowing that $\lambda n T_2\to \chi^2_{n-1}$ (this is starting to not making sense because if it is not minimal it can't be complete, but we can suppose it)

It seems like I have to use Lehmann-Scheffé Theorem proving that $\lambda n T_2$ is unbiased and after that calculate $\operatorname{E}[T_2\mid n\lambda T_2]$. But I don't know how to prove that $\lambda n T_2$ is unbiased. My attempt is:

$$\operatorname{E}[\lambda n T_2] = \int_0^\infty \lambda n T_2 \cdot f_{n-1}(x) dx = \frac{\lambda}{\mu^2 2^{(x-1)/2}\Gamma((n-1)/2)}\int_0^\infty \sum \frac{(x_i-\mu)^2}{x_i} x^{n-1}e^{-x/2} dx$$ Edit 1: As whuber said, this notation doesn't make sense. I was trying to use the definition of expectation:

$$E[g(x)]=\int_\infty^\infty g(x)f(x)dx$$

Edit 2: I've been trying to continue. But I get a solution without using the premise that $T_2$ is complete, so I'm still not sure about my solution. I do this:

We know that $\lambda n T_2\to\chi^2_{n-1}$, so $\operatorname{E}[\lambda n T_2] = n-1$, then

$$\operatorname{E}[\frac{n}{n-1}T_2] = \frac{1}{\lambda}$$

thus

$$S(x_1,\ldots,x_n) = \frac{n}{n-1}T_2 = \frac{1}{n-1}\sum\frac{(x_i-\mu)^2}{x_1}$$

is unbiased for $1/\lambda$ and it is a function of the sufficient statistic T_2. Then $S(x_1,\ldots,x_n)$ is the UMVUE for $1/\lambda$

Oviously I need some help with this. If it is considered two questions and I should split it just let me know and I'll edit this post.

Thanks!

• Due to inconsistencies in your notation (for instance, $f_{n-1}$ is initially defined as a function of $n$ variables but then you appear to apply it to single Real arguments), I cannot comment on whether the integral is appropriate for your problem; but--as written--it is simple to evaluate, since everything involving $x_i$ and $\mu$ can be factored out (being functionally independent of $x$), leaving an integral that obviously is equal to a power of $2$ times a factorial. – whuber May 6 '15 at 0:20
• Thanks, @whuber, I'm a bit confused at this point. I guess I should call $Z=\lambda n T_2$ and say $Z\to\chi^2_{n-1}$ and then calculate the expectation of $Z$? It would be $E[Z]=n-1$, right? I don't know how to continue from here either. I'll edit the question though – Danowsky May 6 '15 at 0:29

The distribution you consider is an Inverse Gaussian distribution. As such, it belongs to exponential families distributions, hence enjoys a natural minimal sufficient statistic that you can derive from expanding the quadratic term in the exponential to recover the natural representation of exponential families: $$f_{\lambda,\mu}(x)=h(x)\exp\{T(\lambda,\mu)\cdot S(x)-\psi(\lambda,\mu)\}$$ (Hint: the detailed answer is provided in the Wikipedia article. Including the MLE of $(\lambda,\mu)$.)
In your question, you ask about a statistic being sufficient for a single parameter: sufficiency is only defined in terms of the global parameter $(\lambda,\mu)$. Since $\sum_i S(x_i)$ is sufficient, you can derive UMVEs for all quantities that allow for unbiased estimators.