$E[\hat{\theta}_{MME}] = E[\frac{1- 2\overline{y}}{\overline{y}-1}] = \int_0^1 \frac{1- 2\overline{y}}{\overline{y}-1}(\theta+1)y^\theta dy$..?

Question

Let $Y_1, Y_2,\dots , Y_n$ denote a random sample from the probability density function

$$f (y | θ)=\begin{cases} (θ + 1)y^θ, & 0 < y < 1; θ > −1,\\ 0 ,& \text{elsewhere}.\end{cases}$$

I found the $\theta$ by using Method of Moment (MME)

$$ \begin{align} \hat{\theta}&=\frac{1- 2\overline{y}}{\overline{y}-1}\\ \end{align} $$

Then I want to find

$$E[\hat{\theta}_{MME}] = E[\frac{1- 2\overline{y}}{\overline{y}-1}] = \int_0^1 \frac{1- 2\overline{y}}{\overline{y}-1}(\theta+1)y^\theta dy$$...How do I go from here since the y are different?

Answer 1

When the $Y_i$s are independant and with distribution density $f$, then for every $F$, $E[F(Y_1, \dots , Y_n)] = \int dy_1 f(y_1) \dots \int dy_n f(y_n) F( y_1, \dots , y_n)$.

Here this is just an application with $F( y_1, \dots , y_n) = \frac 1n (y_1+ \cdots + y_n)$.

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The integral is wrong. It should be

$$\int dy_1 \dots \int dy_n (\theta+1)^n (y_1\dots y_n)^\theta \frac {n-2(y_1+\dots +y_n)}{y_1+\dots +y_n - n} $$