Obtaining the distribution of a test statistic under $H_0$

Question

I start with a sampling distribution of an unknown discrete probability distribution with PMF $g(x)$. I want to test if this distribution is in fact a geometric distribution, this is my $H_0$. Furthermore I have good reason te believe that "checking for" $$\hat{g}(1)(1-\hat{g}(1))-\hat{g}(2)=0$$ is a good criterion to do this in my case.

(Note that you obtain $p(1-p)-p(1-p)=0$ if $g(x)$ corresponds to a geometric distribution)

So I want to build a test statistic, using this criterion. And here starts my question:

1. I have never encountered a test statistic like this, and I want to learn about the art of creating test statistics. When I browse the internet looking for things like this, I encounter the classical test statistics which I do know, but I want to learn more about the techniques and ideas behind the creation of a test statistic.
2. In this case, starting from $\hat{g}(1)(1-\hat{g}(1))-\hat{g}(2)$, how to proceed? How can I even derive the distribution of $\hat{g}(1)$ and $\hat{g}(2)$? In each case the test statistic itself will, under $H_0$, be centered around $0$. My estimators for $\hat{g}(1)$ and $\hat{g}(2)$ are MLE and my sample size is really big, so intuitively I would say that my test statistic will be approximately (?) normally distributed with mean $0$.

Thanks for your input!

Answer 1

I think the following is related to what you are asking:

If you have a i.i.d. sample size $n$ from a distribution where $\mathbb P(X_i=1)=p$ and $\mathbb P(X_i=2)=(1-p)p$, and you observe $X_i=1$ $A$ times and $X_i=2$ $B$ times, then $\mathbb E\left[\frac An \right] = p$ and $\mathbb E\left[\frac Bn \right] = (1-p)p$,

so $\mathbb E\left[\frac An \right]\left(1-\mathbb E\left[\frac An \right]\right) - \mathbb E\left[\frac Bn \right]=0$,

but $\mathbb E\left[\frac An \left(1- \frac An\right)- \frac Bn\right] = -\frac{(1-p)p}{n} \not = 0$. It does tend towards $0$ as $n$ increases.

The distribution of $\frac An \left(1- \frac An\right)- \frac Bn$ is not normal or even close to normal for small $n$: it is a discrete asymmetric distribution. As $n$ increases, it converges in distribution to $0$; empirically, suitably rescaled it appears to converge in distribution towards a normal distribution with mean $0$ as $n$ becomes large.