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Let $Y_1, Y_2,...,Y_n$ be a random sample from a Gamma(2, $\beta$) distribution with pdf

$$f(y) = \frac{y e^{-y/\beta}}{\beta^2} \;, \quad y>0$$

If the maximum likelihood estimator for $\beta$ is:

$$\hat{\beta} = \frac{1}{2n}\sum_\limits{i=1}^{n}Y_i$$ a) Show that $\hat{\beta}$ is an unbiased estimator for $\beta$.
b) Show that $\hat{\beta}$ is a consistent estimator for $\beta$.

Does anyone know how to do this?

I know the bias of an estimator is

$$\hat{\beta} = E[\hat{\beta}] - \beta$$

Adding this after help from @FoobazJohn and reading
Show consistent by satisfying the two limits below: $$\lim_{n \to \infty}E[\hat{\beta}] = \beta \\ \lim_{n \to \infty}Var[\hat{\beta}] = 0 \\ E[\hat{\beta}] = \frac{1}{2n}E\bigg[\sum_\limits{i=1}^{n}Y_i\bigg] = \frac{n2\beta}{2n} = \beta \\ \lim_{n \to \infty}E[\hat{\beta}] = \lim_{n \to \infty}E[\beta] = \beta \\ Var{\hat{\beta}} = \frac{1}{4n^2}Var\bigg[\sum_\limits{i=1}^{n}Y_i\bigg] = \frac{n2\beta^2}{4n^2}=\frac{\beta^2}{2n} \\ \lim_{n \to \infty}Var[\hat{\beta}] = \lim_{n \to \infty}Var[\frac{\beta^2}{2n}] = 0$$

Satisfies consistency.

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  • $\begingroup$ What do you have to do to prove a) and b) ? Could you share your thoughts? $\endgroup$ Commented Oct 6, 2018 at 16:34
  • $\begingroup$ I will get back to you after my exam @StubbornAtom I have not got time for mini tests today. $\endgroup$
    – Bucephalus
    Commented Oct 6, 2018 at 16:43
  • $\begingroup$ You don't have to get back to me or anything. Just make sure you add sufficient context in your posts which would only increase your chances of getting a favourable response. $\endgroup$ Commented Oct 6, 2018 at 16:46
  • $\begingroup$ This is a past exam question. I have so much to learn in the next 24 hours and I was hoping to lean on the community to learn some things while I'm working on others. Not that I'm lazy, just outsourcing to try and get through everything in limited time. As you know, most of my questions have effort in them, but if you see one like this, it is because I have nominated it to get an answer from the community instead of spending an hour learning the stuff. I'm hoping my effort in my past questions would give me some currency to do that, but if you don't think so, I understand. @StubbornAtom $\endgroup$
    – Bucephalus
    Commented Oct 6, 2018 at 17:17
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    $\begingroup$ $E(\hat\beta)\to \beta$ and $\operatorname{Var}(\hat\beta)\to 0$ as $n\to\infty$ is indeed a sufficient condition for consistency of $\hat\beta$. There are some mistakes in the writing but the idea is correct. $\endgroup$ Commented Oct 7, 2018 at 13:49

1 Answer 1

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To show that the estimate is unbiased we have to show that $E\hat{\beta}=\beta$. Since the $Y_i$ are identically distributed and $EY_1=2\beta$, it follows that $E\hat{\beta}=(2n)^{-1}\times n\times 2\beta=\beta$ as desired. To show that it is a consistent estimator one can use the strong law of large numbers to deduce that $$ \hat{\beta}=\frac{1}{2}\times\bar{Y}_n\to\frac{1}{2}EY_1=\beta $$ a.s as $n\to \infty$ as desired.

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  • $\begingroup$ I don't really understand the consistency part John. Can you break it down a little bit more? $\endgroup$
    – Bucephalus
    Commented Oct 6, 2018 at 16:01

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