by Math Avengers
Last Updated December 06, 2018 15:19 PM

I have a test statistics $S(\theta_0) = $ number of $[X_i>0] $ that follows a binomail distribution iwth $p=\frac{1}{2}$. With the standardized test statitics is $S=\frac{S(\theta_0)-(\frac{n}{2})}{\frac{\sqrt(n)}{2}}$, the solution shows that the moment generating function is $M_S(t) = [e^{-(t/2)/(\sqrt{n}/2)}*(\frac{1}{2}e^{t/(\sqrt{n}/2}+\frac{1}{2})]^n$.

My question is, shouldn't it simply be $[\frac{1}{2}*e^{t}+(1-p)]^n$ ?

You do not get the Binomial MGF, since your standardised statistic is not binomial but something else. When you subtract

$S(\theta_0) - \frac{n}{2}$

you are shifting the binomial random variable $S(\theta_0)$ values to the left.

However, binomial random variable is always positive, while your shifted statistic can become negative, so it is not binomial anymore.

Updated January 30, 2018 01:19 AM

Updated October 11, 2017 20:19 PM

- Serverfault Query
- Superuser Query
- Ubuntu Query
- Webapps Query
- Webmasters Query
- Programmers Query
- Dba Query
- Drupal Query
- Wordpress Query
- Magento Query
- Joomla Query
- Android Query
- Apple Query
- Game Query
- Gaming Query
- Blender Query
- Ux Query
- Cooking Query
- Photo Query
- Stats Query
- Math Query
- Diy Query
- Gis Query
- Tex Query
- Meta Query
- Electronics Query
- Stackoverflow Query
- Bitcoin Query
- Ethereum Query