 # Confidence Intervals for an Exponential Distribution. y_1 is distributed f_Y(y∣theta)=thetae^{-theta y} I_{(0,infty)}(y), where theta>0. Analyze the confidence interval for 1/theta given by [L(Y),U(Y)]=[Y,2Y]. 1: Determine the confidence coefficient of this interval. geneth1u 2022-09-14 Answered
Confidence Intervals for an Exponential Distribution
${y}_{1}$ is distributed ${f}_{Y}\left(y\mid \theta \right)=\theta {e}^{-\theta y}{I}_{\left(0,\mathrm{\infty }\right)}\left(y\right)$, where $\theta >0$. Analyze the confidence interval for $\frac{1}{\theta }$ given by $\left[L\left(Y\right),U\left(Y\right)\right]=\left[Y,2Y\right]$
1: Determine the confidence coefficient of this interval.
Thoughts: I realized that my single data point is distributed
$\mathit{\text{exponential}}\left(\beta =\frac{1}{\theta }\right)=\mathit{\text{gamma}}\left(\alpha =1,\beta =\frac{1}{\theta }\right)$. So I did the following steps to get my confidence coefficient:

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

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.
Thoughts: I let $W={Y}^{\frac{1}{\theta }}$, and I performed the transformation to get the distribution of W, and I ended with:

, 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.
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Step 1
(1.) Done, "confidently"...
(2.) The length of the interval $\left[L\left(Y\right),U\left(Y\right)\right]=\left[Y,2Y\right]$ is $2Y-Y=Y$ hence the expected length of the interval is ${\mathbb{E}}_{\theta }\left(Y\right)=1/\theta$.
(3.) The interval $\left[aY,bY\right]$ is a confidence interval for $1/\theta$ with confidence level
${\mathbb{P}}_{\theta }\left(aY⩽1/\theta ⩽bY\right)={\mathbb{P}}_{\theta }\left(1/\left(b\theta \right)⩽Y⩽1/\left(a\theta \right)\right)={\mathrm{e}}^{-1/b}-{\mathrm{e}}^{-1/a}.$
Hence every interval $\left[aY,bY\right]$ such that ${\mathrm{e}}^{-1/b}-{\mathrm{e}}^{-1/a}={\mathrm{e}}^{-1/2}-{\mathrm{e}}^{-1}$ and $b-a<1$ is solution.
Step 2
For example, $a=.5$ yields $b=1.01674$ hence $b-a=.51674<1$ (the minimum length is around $b-a=.471$ for $a=.332$ and $b=.804$).

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