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I'm having trouble with homework and hope somebody can help.

The lifespan of a lightbulb is assumed to be a random variable $X$ with density function: $$f_X(x)=e^{-x/\theta}/\theta,\, 0\leq x $$

The lifespan of a single lightbulb have been measured to 1000 hours. Construct a confidence interval that contains the true values of $\theta$ with $95$% certainty.

My attempt:

$$f_X(x)=e^{-x/\theta}/\theta\implies F_X(t)=1-e^{-t/\theta}$$

So $$F_{X/\theta}(t)=1-e^{-t}$$ is independent of $\theta$

Now $$P(X/c_2\leq \theta <X/c_1)=P(c_1\leq X/\theta < c_2)=F_{X/\theta}(c_2)-F_{X/\theta}(c_1)=(1-e^{-c_2}) - (1-e^{-c_1})=e^{-c_1}-e^{-c_2}$$

And here I am stuck with no way forward. I dont really understand the general procedure here and would really apreciate some help

Regards, Tobias

user25470
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1 Answers1

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The inequality $\mathrm e^{-c_1}-\mathrm e^{-c_2}\geqslant95\%$ is achieved by many $(c_1,c_2)$, for example if $\mathrm e^{-c_1}\geqslant97.5\%$ and $\mathrm e^{-c_2}\leqslant2.5\%$, say $c_1=0.0253$ and $c_2=3.6889$.

Another valid, but asymetric, choice is $c_1=0$, $c_2=2.9958$ (this yields the shortest possible interval). Still another one is $c_1=0.1053$, $c_2=+\infty$.

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