Show that $\hat\theta=\frac{2 \bar Y- 1}{1- \bar Y}$ is a consistent estimator for $\theta$ Let $Y_1,Y_2,...,Y_n$ denote a random sample from the probability density function
$$f(y| \theta)= \begin{cases} ( \theta +1)y^{ \theta}, & 0 < y<1 , \theta> -1 \\ 0, & \mbox{elsewhere},   \end{cases}$$
Find an Estimator for $\theta$ by using the method of moments and show that it is consistent.
I have found the estimator but unsure how to show that it is consistent.
$\mathbb{E}Y=\frac{\theta +1}{\theta +2}$ and $m_1'(u)= \frac{1}{n} \sum_{i=1}^{n}Y_i= \bar Y$
Now, 
$$\mathbb{E}Y=\frac{ \theta +1}{ \theta +2}=m_1'(u)= \frac{1}{n} \sum_{i=1}^{n}Y_i= \bar Y$$
So $$\bar{Y}=\frac{ \theta +1}{ \theta +2} \to \hat{\theta}=\frac{2 \bar Y- 1}{1- \bar Y} $$
Now I am unsure how to show that $\hat\theta=\frac{2 \bar Y- 1}{1- \bar Y}$ is a consistent estimator for $\theta$
Can someone please guide me?
 A: The sequence of estimators $(\hat\theta_n)_n$ is consistent if $\hat\theta_n\to\theta$ in probability with respect to $\mathbb P_\theta$, for every $\theta$. Here, the strong law of large numbers for i.i.d. sequences of integrable random variables implies that $\bar Y_n\to\mathbb E_\theta(Y)=(\theta+1)/(\theta+2)$ almost surely with respect to $\mathbb P_\theta$, hence $\bar Y_n\ne1$ for every $n$ large enough, almost surely, and $\hat\theta_n=(2\bar Y_n-1)/(1-\bar Y_n)$ is well defined for every $n$ large enough, almost surely. Finally, $\hat\theta_n\to(2\mathbb E_\theta(Y)-1)/(1-\mathbb E_\theta(Y))=\theta$ almost surely with respect to $\mathbb P_\theta$. Almost sure convergence implies convergence in probability hence $(\hat\theta_n)_n$ is consistent.
This applies to every i.i.d. random sample with $\mathbb E_\theta(Y)=m(\theta)$, for some homeomorphism $m$.
A: Your distribution is a Beta distribution with parameters $\theta+1$ and $1$, that is
$$
Y \sim \text{Beta}(\theta+1,1).
$$
Consider the case $n=1$, so that $\hat{\theta} = \frac{2Y - 1}{1-Y} = \frac{1}{1-Y} - 2$. Then
$$
E[\frac{1}{1-Y}] = (\theta+1) \int_0^1 (1-y)^{-1} y^\theta dy 
$$
and the integral apparently is unbounded. That is $E[\hat{\theta}] = \infty$ if you will.
