# 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?

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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 $m$.

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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.

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If $E[ \hat \theta]= \infty$ then how is it an unbiased estimator? – math101 Feb 14 '13 at 5:17
It doesn't seem that it is. If you correctly transcribed all the information, it may be that the person asking the question is mistaken. – Glen_b Feb 14 '13 at 5:23
Ohhh shit. The question states that we must show that the estimator is consistent. Sorry my bad – math101 Feb 14 '13 at 5:25
Well, it depends on whether you have the tools to show the result you need from there. It can be done this way, or there are other ways. Do you know Slutsky's theorem? If not, do you know the Chebyshev and Markov inequalities? – Glen_b Feb 14 '13 at 5:40
@Glen_b None of these is useful here. – Did Feb 14 '13 at 6:49