Would someone be able to check my work on this question:
Suppose that $Y_1,Y_2,...,Y_5$ is a random sample from a uniform distribution over the interval (0, theta). Determine if the followng estimators for theta are unbiased or biased
a) $\theta = \dfrac{2}{5}(Y_1 + Y_2 +Y_3 + Y_4 + Y_5)$
b) $\theta = Y_{(5)}$, the fifth order statistic
a) $\theta = \dfrac{2}{5}(Y_1 + Y_2 +Y_3 + Y_4 + Y_5)$ $$E(\theta) = E(\dfrac{2}{5}(Y_1 + Y_2 +Y_3 + Y_4 + Y_5))$$ $$ = \dfrac{2}{5}E((Y_1 + Y_2 +Y_3 + Y_4 + Y_5))$$ $$ = \dfrac{2}{5}(E(Y_1) + E(Y_2) + E(Y_3) + E(Y_4) + E(Y_5))$$ $$E(Y_i) = \dfrac{0+\theta}{2} = \dfrac{\theta}{2}$$ $$E(\theta) = \dfrac{2}{5}(\dfrac{\theta}{2} + \dfrac{\theta}{2} + \dfrac{\theta}{2} + \dfrac{\theta}{2} + \dfrac{\theta}{2}) = \dfrac{2}{5}(\dfrac{5 \theta}{2}) = \theta$$
Thus, the estimator is biased
b)
$$g_(5) = n[F(y)]^{n-1} f(y)$$
$$F(y) = \int_{0}^{y} \dfrac{1}{\theta} dt = \dfrac{y}{\theta}$$ $$g_(5) = 5[\frac{y}{\theta}]^4 (\dfrac{1}{\theta}) = \dfrac{5y^4}{\theta^5}$$ $$E(\theta) = E(\dfrac{5y^4}{\theta^5}) = y*\dfrac{5y^4}{\theta ^5} = \dfrac{5y^5}{\theta ^5}$$
Thus, the estimator is biased
I felt really uncertain of my work, I really appreciate any help! Thanks in advanced!
Your calculation of (a) is correct, but your conclusion should say "unbiased," since you found that the expectation equals the parameter.
Regarding (b), I would proceed from more elementary principles: $$\begin{align*} F_{Y_{(5)}}(y) &= \Pr[Y_{(5)} \le y] \\ &= \prod_{i=1}^5 \Pr[Y_i \le y] \\ &= (F_{Y}(y))^5 \\ &= (y/\theta)^5, \end{align*}$$ hence $$f_{Y_{(5)}}(y) = 5y^4/\theta^5, \quad 0 \le y \le \theta.$$ Up to this point, your solution is correct. However, it follows that $$\operatorname{E}[Y_{(5)}] = \int_{y=0}^\theta y \cdot \frac{5y^4}{\theta^5} \, dy = \frac{5}{\theta^5} \left[ \frac{y^6}{6} \right]_{y=0}^\theta = \frac{5}{6}\theta.$$ Therefore this estimator is biased. You can also see how to modify this estimator to make it unbiased in the general case of a sample size of $n \ge 1$; e.g., you could choose $$\hat\theta = \frac{n+1}{n} Y_{(n)}.$$