Following my previous query, I have decided to give a full blown version of it and my approach and how far I got. Problem is my notes aren't very useful. I have one last bit I just don't understand due to its ambiguity.
A particle moves along a line with uniform velocity $v$ which is unknown. The time taken to cover a known distance is $d$ and recorded or $n$ independent experiments giving $y_1,y_2,...y_n$ observations. $y_i$ are assumed to be realizations of the random variable $Y$ with PDF
$$f(y;\mu)=\frac{1}{\sqrt{2 \pi y^3}}e^{-\frac{(y-\mu)^2}{2 \mu ^2 y}}$$
where $y>0$ and $\mu= \frac{d}{v}$. Write down the log likelihood function and show that the MLE $\bar{\mu}=\bar{y}$. Using known properties of the score, show that $\bar{\mu}$ is unbiased and has asymptotic variance $\frac{\mu^3}{n}$
The following is what I need help on
Describe the test $H_0:\mu=\mu_0$ using the asymptotic normal distribution of the score. Obtain the asymptotic likelihood ratio test of the same hypothesis and show that the two tests are approximately equivalent.
"approximately"?"Equivalent"? What exactly is the definition of two tests being equivalent? And to what extent can I measure "approximately"?
Here are the work I have done for the first bit of the question and the bit in question.($\bar{\mu}$ is the MLE and $\mu$ is the true value in the following)
Likelihood function:
$$L(\mu)=\frac{1}{\sqrt{(2 \pi)^n \Pi y_i^3}}e^{- \sum \frac{(y_i-\mu)^2}{2 \mu^2 y_i}}$$
log likelihood function:
$$l(\mu)=-\frac{n}{2}log2\pi - \frac{1}{2}log\Pi y_i^3 - \sum \frac{(y_i- \mu)^2}{2 \mu ^2 y_i}$$
Score function($:= l'(\mu)$):
$$U=- \sum \frac{\mu - y_i}{\mu ^3}$$
Fisher information:
$$I(\mu)=\frac{n}{\mu ^3}$$
The asymptotic normal distribution statistic (according to my notes, this is $\frac{U(\mu)}{\sqrt{I(\mu)}}$)
$$\frac{U(\mu)}{\sqrt{I(\mu)}}=-\sqrt{\frac{1}{n \mu^3}}\big(n \mu - \sum y_i\big)$$
The likelihood ratio(my notes says this is $\frac{L(\mu)}{L(\bar{\mu})}$ but this one has been taken with $log$ ):
$$\frac{L(\mu)}{L(\bar{\mu})}=\frac{(\mu^2- \bar{\mu}^2)}{2 \mu \bar{\mu}}\sum y_i- n \frac{\bar{\mu}-\mu}{\bar{\mu} \mu}$$
Test statistic for likelihood ratio test is $-2log\frac{L(\mu)}{L(\bar{\mu})}$ according to my notes again.
Some facts
$$\mathbb{E}[\bar{\mu}]=\mu, \bar{\mu}=\bar{y}$$
This is as far as I have. Am I to show that $\frac{L(\mu)}{L(\bar{\mu})}=\frac{(\mu^2- \bar{\mu}^2)}{2 \mu \bar{\mu}}\sum y_i- n \frac{\bar{\mu}-\mu}{\bar{\mu} \mu}$ is approximately equal to $\frac{U(\mu)}{\sqrt{I(\mu)}}=-\sqrt{\frac{1}{n \mu^3}}\big(n \mu - \sum y_i\big)$? If so, it beats me, they have some similarities but far from equivalent.
Does anyone know what I am to do with all of these bunch of equations? I have no clue