Let $Y_1, Y_2, . . . , Y_n$ denote a random sample from the probability density function
$$f (y \mid \theta)=\begin{cases} (\theta + 1)y^\theta, & 0 < y < 1; \theta > −1,\\ 0 ,& \text{ elsewhere }\end{cases}$$
Find an estimator for $\theta$ by the method of moments.
I am told that that $μ = \frac{\theta + 1}{\theta + 2} $. I am wondering how do they show this ?
Best Answer
The motivation of the method of moments estimate is that it produces a model that has the same first $n$ raw moments as the data (as represented by the empirical distribution). Let $Y_1, Y_2, \cdots, Y_n$ be a random sample, therefore $$ \text{E}\left[Y^n\right]=\frac{1}{n}\sum_{i=1}^n y_i^n.\tag1 $$ Let us obtain the first raw moment of the data. $$ \overline{y}=\frac{1}{n}\sum_{i=1}^n y_i.\tag2 $$ Now, we obtain the first raw moment of the sample distribution. $$ \begin{align} \mu&=\text{E}[Y]\\ &=\int_{y=0}^1 yf(y|\theta)\ dy\\ &=\int_{y=0}^1 y(\theta+1)y^\theta\ dy\\ &=(\theta+1)\int_{y=0}^1 y^{\theta+1}\ dy\\ &=(\theta+1)\cdot\left.\frac{y^{\theta+2}}{\theta+2}\right|_{y=0}^1\\ &=\frac{\theta+1}{\theta+2}.\tag3 \end{align} $$ Thus, based on $(1)$, from $(2)$ and $(3)$ we obtain $$ \begin{align} \frac{\hat{\theta}+1}{\hat{\theta}+2}&=\overline{y}\\ \hat{\theta}+1&=\overline{y}(\hat{\theta}+2)\\ \hat{\theta}+1&=\overline{y}\hat{\theta}+2\overline{y}\\ \hat{\theta}-\overline{y}\hat{\theta}&=2\overline{y}-1\\ \hat{\theta}(1-\overline{y})&=2\overline{y}-1\\ \hat{\theta}&=\frac{2\overline{y}-1}{1-\overline{y}}\\ &=-\frac{2\overline{y}-1}{\overline{y}-1}. \end{align} $$
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