Let $X_1, X_2, ..., X_n$ be a random sample from a distribution who's PDF is given by $f(x; \theta)=(\theta+1)x^{\theta} $ for $0 \leq x \leq 1$ or $0$ otherwise.

Find the method-of-moments estimator for $\theta$.

So I have done the following:

$\mathbb{E}(X)=\int_{-\infty}^{\infty} xf(x) dx = (\theta +1)\int_{0}^{1}x^{\theta+1}dx=\frac{\theta +1}{\theta+2}$

I am unsure now how to equate this to $\bar{X}$

  • 1
    $\begingroup$ Seems like all you need to do is just to equate it to the sample mean $\frac{\theta + 1}{\theta +2} = \frac{1}{n}\sum_{n=1}^n X_i$ and solve the equation for $\theta$ (the obtained value will be the estimator $\hat{\theta}$) $\endgroup$ – applyb Aug 25 '16 at 19:17
  • $\begingroup$ You simply equate corresponding population and sample moments to obtain the estimator for $\theta$. $\endgroup$ – Glen_b Aug 31 '16 at 3:01

It's like Beck says. $\theta +1=\bar X (\theta+2) \Leftrightarrow \theta=\frac{2\bar X -1}{1-\bar X}$

  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ – Henrik Aug 25 '16 at 20:07
  • $\begingroup$ @Henrik how come? the estimator for theta by estimator of moments it the one in the answer... $\endgroup$ – An old man in the sea. Aug 25 '16 at 20:51

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