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Let $Y_1, Y_2,\ldots,Y_n$ denote a random sample from the uniform distribution on the interval $(θ, θ + 1)$. Let $$ \hat{\theta}_2 = Y_{(n)} - \frac{n}{n+1}$$

Find the efficiency of $θ^1$ relative to $θ^2$

We have $Y_i\sim\mathcal{U}(\theta,\theta+1)$ and CDF of $Y_i$ based on Wikipedia $$ G_{Y_i}(y)=\Pr[Y_i\le y]=\frac{y-\theta}{\theta+1-\theta}=y-\theta. $$ Here, $Y_{(n)}$ is $n$-th order statistics. Therefore, $Y_{(n)}=\max[Y_1,\ldots, Y_n]$. Note that $Y_{(n)}\le y$ equivalence to $Y_i\le y$ for $i=1,2,\ldots,n$. Hence, for $\theta< y<\theta+1$, the fact that $Y_1,Y_2,\ldots, Y_n$ are i.i.d. implies $$ G_{Y_{(n)}}(y)=\Pr[Y_{(n)}\le y]=\Pr[Y_1\le y,Y_2\le y,\ldots, Y_n\le y]=(\Pr[Y_i\le y])^n=\left(y-\theta\right)^{n}. $$ The PDF of $Y_{(n)}$ is $$ g_{Y_{(n)}}(y)=\frac{d}{dy}G_{Y_{(n)}}(y)=\frac{d}{dy}(y-\theta)^n=n(y-\theta)^{n-1}. $$ The expected value of $Y_{(n)}$ is $$ \begin{align} \text{E}\left[Y_{(n)}\right]&=\int_{y=\theta}^{\theta+1}yg_{Y_{(n)}}(y)\ dy\\ &=\int_{y=\theta}^{\theta+1}yn(y-\theta)^{n-1}\ dy\\ &=n\int_{y=\theta}^{\theta+1}y(y-\theta)^{n-1}\ dy. \end{align} $$

I have trouble finding $$ \text{Var}\left[\hat{\theta}_{2}\right]=\text{Var}\left[Y_{(n)}-\frac{n}{n+1}\right]=\text{Var}\left[Y_{(n)}\right]=\text{E}\left[Y_{(n)}^2\right]-\left(\text{E}\left[Y_{(n)}\right]\right)^2. $$

I found that $$ \begin{align} \text{E}\left[Y_{(n)}\right]&=n\left[\frac{y(y-\theta)^n}{n+1}-\frac{\theta(y-\theta)^n}{n(n+1)}\right]_{y=\theta}^{\theta+1}\\ &=\frac{n(\theta+1)}{n+1}+\frac{\theta}{n+1}\\ &=\theta+\frac{n}{n+1}. \end{align} $$

The way I calculate $E(Y_{(n)}^2)$ is the following:

$$E(Y_{(n)}^2) = ny^2(y-\theta)^{n-1} = n\left[\left.y^2\frac{(y-\theta)^n}{n} \right|_\theta^{\theta+1} - \frac{2}{n} \int_\theta^{\theta+1} y(y-\theta)^n \,dy\right]$$

$$= \left.(\theta+1)^2 - 2\left(y\frac{(y-\theta)^{n+1}}{n+1} \right|_\theta^{\theta+1} - \int_\theta^{\theta+1} \frac{(y-\theta)^{n+1}}{n+1} dy\right) = (\theta+1)^2 -2 \left(\frac{\theta+1}{n+1} - \left.\frac{(y-\theta)^{n+2}}{(n+1)(n+2)}\right|_\theta^{\theta+1}\right)$$

$$= (\theta+1)^2- 2\frac{\theta +1}{n+1} - \frac{1}{(n+1)(n+2)}$$

Then I use $E(Y_{(n)}^2) - E(Y_{(n)})^2$, based on wolframalpha which gives me the following result ...please go here ...which is different from the correct answer $\text {Var} [\hat{\theta}_2]= V(Y(n))=\frac{n}{(n+2)(n+1)^2}$....Could anyone please check why?

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  • $\begingroup$ I already tried what I knew .... I just wanted to see if someone can help spot my mistake... $\endgroup$ – afsdf dfsaf Apr 12 '14 at 14:43
  • $\begingroup$ Try to fix the formula. What is $Y_{(n)}$? $\endgroup$ – leonbloy Apr 12 '14 at 14:44
  • $\begingroup$ If it is appropriate for you to be seeking such help, then please ask your tutor: then they will understand what part you have been able to answer yourself, and be able to mark accordingly. $\endgroup$ – wolfies Apr 12 '14 at 14:45
  • $\begingroup$ @leonbloy: I tried my best to fix the formula...but it still looks the way it does...but you could focus on the answer...I just edited my questions to see if that address your concern on $Y_{(n)}$ $\endgroup$ – afsdf dfsaf Apr 12 '14 at 14:47
  • $\begingroup$ @leonbloy: just fix the formule $\endgroup$ – afsdf dfsaf Apr 12 '14 at 15:05
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Variance can be evaluated as follows $$ \text{Var}\left[\hat{\theta}_{2}\right]=\text{Var}\left[Y_{(n)}-\frac{n}{n+1}\right]=\text{Var}\left[Y_{(n)}\right]=\text{E}\left[Y_{(n)}^2\right]-\left(\text{E}\left[Y_{(n)}\right]\right)^2. $$ First, we calculate $\text{E}\left[Y_{(n)}^2\right]$. Using the result from here, we obtain $$ \text{E}\left[Y_{(n)}^2\right]=n\int_{y=\theta}^{\theta+1}y^2(y-\theta)^{n-1}\ dy. $$ Using IBP, let $u=y^2\Rightarrow du=2y\ dy$ and $dv=(y-\theta)^{n-1}\ dy\Rightarrow v=\dfrac{(y-\theta)^n}{n}$. Hence $$ \begin{align} n\int_{y=\theta}^{\theta+1}y^2(y-\theta)^{n-1}\ dy&=n\left[\left.\dfrac{y^2(y-\theta)^n}{n}\right|_{y=\theta}^{\theta+1}-\int_{y=\theta}^{\theta+1}\dfrac{(y-\theta)^n}{n} 2y\ dy\right]\\ &=(\theta+1)^2-2\int_{y=\theta}^{\theta+1}y(y-\theta)^{n}\ dy\;\;\;\Rightarrow\;\;\;\text{again we use IBP here.}\\ &=(\theta+1)^2-2\left[\left.\dfrac{y(y-\theta)^{n+1}}{n+1}\right|_{y=\theta}^{\theta+1}-\int_{y=\theta}^{\theta+1}\dfrac{(y-\theta)^{n+2}}{n+1} dy\right]\\ &=(\theta+1)^2-2\left[\dfrac{\theta+1}{n+1}-\left.\dfrac{(y-\theta)^{n+2}}{(n+1)(n+2)}\right|_{y=\theta}^{\theta+1}\right]\\ \text{E}\left[Y_{(n)}^2\right]&=(\theta+1)^2-\dfrac{2(\theta+1)}{n+1}+\dfrac{2}{(n+1)(n+2)}. \end{align} $$ You already have $$ \text{E}\left[Y_{(n)}\right]=\theta+\frac{n}{n+1}. $$ Thus $$ \begin{align} \text{Var}\left[\hat{\theta}_{2}\right]&=\text{E}\left[Y_{(n)}^2\right]-\left(\text{E}\left[Y_{(n)}\right]\right)^2\\ &=(\theta+1)^2-\dfrac{2(\theta+1)}{n+1}+\dfrac{2}{(n+1)(n+2)}-\left(\theta+\frac{n}{n+1}\right)^2\\ &=\theta^2+2\theta+1-\dfrac{2(\theta+1)}{n+1}+\dfrac{2}{(n+1)(n+2)}-\theta^2-\frac{2n\theta}{n+1}-\frac{n^2}{(n+1)^2}\\ &=\dfrac{(2\theta+1)(n+1)-2(\theta+1)}{n+1}+\dfrac{2}{(n+1)(n+2)}-\frac{2n\theta}{n+1}-\frac{n^2}{(n+1)^2}\\ &=\dfrac{2n\theta+2\theta+n+1-2\theta-2}{n+1}+\dfrac{2}{(n+1)(n+2)}-\frac{2n\theta}{n+1}-\frac{n^2}{(n+1)^2}\\ &=\dfrac{2n\theta+n-1}{n+1}+\dfrac{2}{(n+1)(n+2)}-\frac{2n\theta}{n+1}-\frac{n^2}{(n+1)^2}\\ &=\dfrac{n-1}{n+1}+\dfrac{2}{(n+1)(n+2)}-\frac{n^2}{(n+1)^2}\\ &=\dfrac{(n-1)(n+1)+2}{(n+1)(n+2)}-\frac{n^2}{(n+1)^2}\\ &=\dfrac{n(n+1)}{(n+1)(n+2)}-\frac{n^2}{(n+1)^2}\\ &=\dfrac{n}{n+2}-\frac{n^2}{(n+1)^2}\\ &=\frac{n(n+1)^2-n^2(n+2)}{(n+2)(n+1)^2}\\ &=\frac{n}{(n+2)(n+1)^2}. \end{align} $$

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$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$

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  • $\begingroup$ I cannot believe it I use my tablet to answer this. I hope this helps. $\endgroup$ – Tunk-Fey Apr 12 '14 at 15:09
  • $\begingroup$ Thanks a lot...I made two minor mistakes.... $\endgroup$ – afsdf dfsaf Apr 12 '14 at 15:17
  • $\begingroup$ If you are interest in looking for a related post, please go to the following link math.stackexchange.com/questions/750358/… $\endgroup$ – afsdf dfsaf Apr 12 '14 at 15:30
  • $\begingroup$ It seems nothing is wrong. Your answer is correct. Are you sure that's the complete question? Maybe you missed something when you were typing it. $\endgroup$ – Tunk-Fey Apr 12 '14 at 15:46
  • $\begingroup$ Let chat on that question, which answer you think is right...please answer in that post $\endgroup$ – afsdf dfsaf Apr 12 '14 at 15:47

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