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!
Best Answer
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 $$\hat\theta = \frac{n+1}{n} Y_{(n)}.$$