An urn contains $\theta$ black balls and $N-\theta$ white balls. A sample of $n$ balls is to be selected without replacement. Let $Y$ denote the number of black balls in the sample. Show that $(N/n)Y$ is the method-of-moments (MOM) estimators of $\theta$.
My approach: I know that:
Method of moment (MOM): Choose as estimates values of the parameters that are solutions of the equations $\mu_{k}'=m_{k}'$, for $k=1,2,\ldots,t$ where $t$ is the number of parameters to be estimated and $\mu_{k}'=\mathbb{E}[Y^{k}]$ is the $k$-th moment of a random variable and $\displaystyle m_{k}'=\frac{1}{n}\sum_{i=1}^{n}Y_{i}^{k}$ is the corresponding $k$-th sample moment.
Now, 1) I need to know of the distribution of random variable $Y$, so I think that our sole observation $Y$ is hypergeometric, but I don't sure about it. How can I check the first part?
- I think here $t=1$, so we need to solve $\mu_{1}'=m_{1}'$. I think that part is easy if a can find the distribution for $Y$.
Best Answer
Considering the Hypergeometric distribution,
$$\mu=\frac{\theta n}{N}$$
thus
$$\theta=\frac{N}{n}\mu$$
thus the MOM estimator of $\theta$ is
$$\hat{\theta}_{MM}=\frac{N}{n}\overline{Y}_n$$
because it has been obtained substituting the first moment with the first sample moment