We all know that the formula of the score test is: $$ S = \frac{\{l'(\theta_0)\}^2}{E\{-l''(\theta)\}\bigg|_{\theta_0}} $$ where $l'$ and $l''$ are $log$ $likelihoods$

What I need to do is represent that score statistic in the form of: $$S = (1-\widehat{\beta})^2\frac{n}{\widehat{\beta}^2}$$

this are the parameters that I have:

$$\theta_0 = 1\\ \theta = \beta\\ E\{-l''(\theta)\}\bigg|_{\theta_0} = n \\ E\{-l''(\beta)\} = \frac{n}{\beta^2}\\ l'(\theta_0) = n-\sum_{i=1}^{n}logX_i \\ \widehat{\beta} = \frac{n}{\sum_{i=1}^{n}logX_i}$$

So in order to re-write S in the for that it is required, my plan of attack was to divide the numerator and the denominator by $\sum_{i=1}^{n}logX_i$

This are the steps that I have taken to try and resolve this issue:

$$S = \frac{\{n-\sum_{i=1}^{n}logX_i\}^2}{n}$$

$$S = \frac{\left\{\frac{n-\sum_{i=1}^{n}logX_i}{\sum_{i=1}^{n}logX_i}\right\}^2} {\frac{n}{\sum_{i=1}^{n}logX_i}}$$

$$S = \frac{\left\{\frac{n}{\sum_{i=1}^{n}logX_i}-\frac{\sum_{i=1}^{n}logX_i}{\sum_{i=1}^{n}logX_i}\right\}^2} {\frac{n}{\sum_{i=1}^{n}logX_i}}$$

$$S = (\widehat{\beta}-1)^2\frac{1}{\widehat{\beta}}$$

So I have a couple of issues here. the first one is that i have got $(\widehat{\beta}-1)^2$ which has the opposite signs that need to make it look in the same form as $S = (1-\widehat{\beta})^2\frac{n}{\widehat{\beta}^2}$ and the second issue is that I have run out of ideas on how to make $\frac{1}{\widehat{\beta}}$ convert to $\frac{n}{\widehat{\beta}^2}$. no idea where the $n$ comes from! Sorry about the long question but i wanted to be as detailed as possible and show that i have been doing my best to solve this problem

  • $\begingroup$ Actually, you divided the denominator by $\sum_{i=1}^{n}logX_i$ and numerator by the $(\sum_{i=1}^{n}logX_i)^2$. You should probably divide both by $(\sum_{i=1}^{n}logX_i)^2$. $\endgroup$ – Matthias Klupsch Jan 9 '20 at 20:58
  • $\begingroup$ Thanks @MatthiasKlupsch . can you also help me understand how do i get the n on the numerator?... i have 0 ideas left! $\endgroup$ – Raul Gonzales Jan 10 '20 at 13:21

$$ S = \frac{\{n-\sum_{i=1}^{n}logX_i\}^2}{n} = \frac{\left\{\frac{n-\sum_{i=1}^{n}logX_i}{\sum_{i=1}^{n}logX_i}\right\}^2} {\frac{n}{(\sum_{i=1}^{n}logX_i)^2}} = \frac{\left\{\frac{n}{\sum_{i=1}^{n}logX_i}-\frac{\sum_{i=1}^{n}logX_i}{\sum_{i=1}^{n}logX_i}\right\}^2} {\frac{n^2}{n (\sum_{i=1}^{n}logX_i)^2}}$$

with the numerator being equal to $(\widehat{\beta} - 1)^2 = (1 - \widehat{\beta})^2$ (note the square) and the denominator is $\frac{\widehat{\beta}^2}{n}$.


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.