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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?

Johnny Westerling
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math101
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2 Answers2

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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 $m$.

Did
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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.

passerby51
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  • If $E[ \hat \theta]= \infty$ then how is it an unbiased estimator? – math101 Feb 14 '13 at 05:17
  • 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 05:23
  • Ohhh shit. The question states that we must show that the estimator is consistent. Sorry my bad – math101 Feb 14 '13 at 05:25
  • An estimator can be biased and consistent too. hmmm – math101 Feb 14 '13 at 05:29
  • @Glen_b So I guess we can say that $\frac{1}{n} \sum_{i=1}^{n}Y_i \to \bar Y \to E(Y)$ Am I on the correct path? – math101 Feb 14 '13 at 05:36
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    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 05:40
  • Yes we kind of surfaced it – math101 Feb 14 '13 at 05:44
  • If its been mentioned, then you should be able to use the division form of Slutsky to get you there. – Glen_b Feb 14 '13 at 05:44
  • Well it has been mentioned but I am not familiar with the method. I do know the Chebyshev and Markov Inequalities but not sure how it fits in with all this – math101 Feb 14 '13 at 05:47
  • I think I will try to play around with Chebyshev's and Markov's Inequality. I am more familiar with that – math101 Feb 14 '13 at 05:58
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    @Glen_b None of these is useful here. – Did Feb 14 '13 at 06:49
  • @Did Why do you say Slutsky isn't useful here? It looks to me like it should work: Let $T_n = 1$ and $S_n = (1-\bar{Y})$ and hence $T_n/S_n \xrightarrow{p} 1/(1-E(Y))$. From there, the result should follow, shouldn't it? Did I miss something? – Glen_b Feb 14 '13 at 07:29
  • @Glen_b Where des one need Slutsky here? The fact that $\bar Y\to E(Y)$ is the LLN, hence $1/(1-\bar Y)$ is also the LLN plus the continuity of the function $y\mapsto1/(1-y)$ on $y\ne1$. – Did Feb 14 '13 at 09:04
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    Well, yes, you're right - in this case Slutsky is more than is needed. But nonetheless, it can be used, so 'none of these is useful' seems too strong. – Glen_b Feb 14 '13 at 14:51