Let $X$ such that $f_{X}(x\mid\theta) = \theta e^{-\theta x} I_{(0, \infty)}(x)$, where $\theta > 0$. If $[X, 2X]$ is a confidence interval for $\frac{1}{\theta}$:

a)Find the confidence coefficient of this interval


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

So the confidence is 0.23865

b) Find other interval with shorter expected length and the same confidence coefficient.

$\bf{Thoughts:}$ The expected length of the confidence interval in a) is $\mathbb E (2X-X)= 1/\theta$.

We can choose other interval $[aY,bY]$ for $1/\theta$, so the expected length would be $(b-a)\frac{1}{\theta}$ then we have to find a, b such that $(b-a) \lt 1$ and $\mathrm e^{-1/b}-\mathrm e^{-1/a}=\mathrm e^{-1/2}-\mathrm e^{-1}$. I can give an arbitrary value to a and find b, but how can I find the values of a and b to minimize the expected length?


This is just a problem in calculus. Set $\alpha = e^{-1/b} - e^{-1/a}$, we find that,

$$e^{-1/b} = \alpha + e^{-1/a} \Rightarrow b = -\frac{1}{\ln(\alpha + e^{-1/a})} $$

So now,

$$f(a) = b - a = \frac{-1}{\ln(\alpha + e^{-1/a})} - a$$

We have to minimise $f$ with respect to the constraint that $f > 0$. If we differentiate, we obtain:

$$f'(a) = \frac{e^{-1/a}}{a^2(\alpha + e^{-1/a})(\ln(\alpha + e^{-1/a}))^2} - 1 $$

A numerical solution puts the minimum at $a \simeq 0.3261$, use that to obtain $b$ to get an approximate minimum.


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