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

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


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

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.


  • 1
    $\begingroup$ 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. $\endgroup$ – whuber May 6 '15 at 0:20
  • $\begingroup$ 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 $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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