# Re-writing the Score Test Statistic

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

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

## 1 Answer

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

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