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$y_{1}$ is distributed $f_{Y}(y\mid\theta) = \theta e^{-\theta y} I_{(0, \infty)}(y)$, where $\theta > 0$. Analyze the confidence interval for $\frac{1}{\theta}$ given by $[L(Y), U(Y)] = [Y, 2Y]$.

1: Determine the confidence coefficient of this interval.

$\bf{Thoughts:}$ I realized that my single data point is distributed $\textit{exponential}(\beta=\frac{1}{\theta}) = \textit{gamma}(\alpha = 1, \beta = \frac{1}{\theta})$. So I did the following steps to get my confidence coefficient:

$$\ P_{\theta}(\frac{1}{\theta}\in [Y, 2Y])\\ = P_{\theta}(Y \leq \frac{1}{\theta} \leq 2Y)\\ = P_{\theta}(\frac{1}{2\theta} \leq Y \leq\frac{1}{\theta})\\ = \int\limits_{\frac{1}{2\theta}}^{\frac{1}{\theta}} \! \theta e^{-\theta y} \mathrm{d}y\\ = e^{-.5} - e^{-1}\\ = .23865$$

So I got .23685 as my confidence coefficient, and I'm fairly confident that that is the correct answer.

2: Determine the expected length of this interval.

$\bf{Thoughts:}$ So I'm not too sure what to do here, but this is what I have written down.

$$\ P_{\theta}(\frac{1}{\theta}\in [Y, 2Y])\\ = P_{\theta}(Y \leq \frac{1}{\theta} \leq 2Y)\\ = P_{\theta}(\frac{1}{2\theta} \leq Y \leq \frac{1}{\theta})\\ = P(e^{-1} \leq Y \leq e^{-.5}) = .23865$$

Since my 'a' and 'b' values were $e^{-1}$ and $e^{-.5}$, respectively. Now I don't know where to go from there to find the expected length. Do I use the gamma probability table? Any help would be greatly appreciated.

3: Find a different confidence interval for $\frac{1}{\theta}$ with the exact same confidence coefficient, but the expected length is smaller than part 2's expected length. Find an interval where the characteristics, same confidence coefficient and expected length is smaller than part 2's, are satisfied.

$\bf{Thoughts:}$ I let $W = Y^{\frac{1}{\theta}}$, and I performed the transformation to get the distribution of W, and I ended with: $$\ f_{W}(w\mid \theta) = \theta^{2} e^{-\theta w^{\theta}}w^{\theta -1}$$, which I think looks sorta like the gamma distribution, but then I don't know what to do from there. Any help would be appreciated. Thanks in advance!

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why do you need to find the distribution of $W=Y^{\frac{1}{\theta}}$? it seems like you are looking for $\mathbb{E}_{1/\theta}Y$. –  Seyhmus Güngören Mar 9 '13 at 3:35
    
Then, I'm not sure what to do for part c, if what I had is not right. Since I'm not sure what to do when trying to find another interval for $\frac{1}{\theta}$. How do you know that I need to look for $E_{\frac{1}{\theta}}[Y]$? –  user61752 Mar 9 '13 at 4:50
    
I can not say $100\%$ that it is the case. But It seems that your interval is $2Y-Y=Y$. If you were dealing with $\theta$ you would have the density and simply you would get $\mathbb{E}_\theta$. Now you are looking for the expected value of $Y$ but when your density is identified in terms of $1/\theta$. –  Seyhmus Güngören Mar 9 '13 at 11:33
    
So then are both parts 2 and 3 looking for $E_{\frac{1}{\theta}}$? –  user61752 Mar 9 '13 at 21:30

1 Answer 1

up vote 0 down vote accepted

(1.) Done, "confidently"...

(2.) The length of the interval $[L(Y),U(Y)]=[Y,2Y]$ is $2Y-Y=Y$ hence the expected length of the interval is $\mathbb E_\theta(Y)=1/\theta$.

(3.) The interval $[aY,bY]$ is a confidence interval for $1/\theta$ with confidence level $$ \mathbb P_\theta(aY\leqslant1/\theta\leqslant bY)=\mathbb P_\theta(1/(b\theta)\leqslant Y\leqslant1/(a\theta))=\mathrm e^{-1/b}-\mathrm e^{-1/a}. $$ Hence every interval $[aY,bY]$ such that $\mathrm e^{-1/b}-\mathrm e^{-1/a}=\mathrm e^{-1/2}-\mathrm e^{-1}$ and $b-a\lt1$ is solution.

For example, $a=.5$ yields $b=1.01674$ hence $b-a=.51674\lt1$ (the minimum length is around $b-a=.471$ for $a=.332$ and $b=.804$).

Find an interval where the characteristics are satisfied.

??

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