Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

1 Answer 1

up vote 0 down vote accepted

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 $c_1=0.1053$, $c_2=+\infty$.

share|improve this answer
    
So it's almost arbitrary? –  user25470 Jan 20 '13 at 19:49
    
Several intervals are valid, yes. –  Did Jan 20 '13 at 19:55
    
Thanks, i susoected as much but it''s rather daunting to just jump to that conclusion –  user25470 Jan 20 '13 at 20:56
    
Not really. Many subintervals of (0,1) are longer than 90% of its length. –  Did Jan 20 '13 at 20:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.