# Finding shortest Confidence Interval for an Exponential Distribution

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

$$\bf{Thoughts:}$$

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

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?

## 1 Answer

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.