1
$\begingroup$

I have a lightbulb that has an exponential lifetime distribution with mean $\mu$ months. So if I construct a pdf $f(x)$ and cdf $F(x)$ with parameter $\lambda$, and since $E[X] = \frac{1}{\lambda}$,

\begin{align} f(x) &= \frac{1}{\mu}e^{-\frac{1}{\mu}x} \\ F(x) &= 1-e^{-\frac{1}{\mu}x} \end{align}

Year $Y$ can be odd if it's the $1$st year ($12$ months), $3$rd year ($36$ months) and so on.

I'm having troubles constructing the $P(Y= \mathrm{odd})$ model here, as I don't know how to put together the infinite odd years described above.

$\endgroup$
  • 2
    $\begingroup$ If $X$ is the random month in which the lightbulb kaputs, and $Y$ is the year of that happening, then $Y$ is odd whenever $X$ is between $1$ and $12$, or $25$ and $36$, or $49$ and $60$, and so on. $\endgroup$ – Rócherz Mar 15 at 1:21
1
$\begingroup$

There's only one parameter in this exponential model – $\lambda=\frac1\mu$. Thus $$F_Y(y)=1-e^{-\lambda x}$$ $$P(12k<Y<12(k+1))=F_Y(12(k+1))-F_Y(12k)=1-e^{-12\lambda(k+1)}-1+e^{-12\lambda k}$$ $$=-e^{-12\lambda(k+1)}+e^{-12\lambda k}=e^{-12\lambda k}(1-e^{-12\lambda})$$ Then the probability the bulb fails in an odd year is an infinite sum with $k=2n=0,2,4\dots$: $$\sum_{n=0}^\infty e^{-24\lambda n}(1-e^{-12\lambda})$$ $$=(1-e^{-12\lambda})\sum_{n=0}^\infty(e^{-24\lambda})^n=\frac{1-e^{-12\lambda}}{1-e^{-24\lambda}}$$ $$=\frac{1-e^{-12/\mu}}{1-e^{-24/\mu}}$$ (We can derive this faster by using the memorylessness of the exponential distribution; in each 24-month cycle the probability of the bulb failing in the first half of that cycle given that it survives to the start of that cycle is constant, and equal to $\frac{F_Y(12)}{F_Y(24)}$.)

$\endgroup$
  • $\begingroup$ So $k$ here has to be even right? $\endgroup$ – PTN Mar 15 at 1:46
  • $\begingroup$ @PTN Yes. ${}{}$ $\endgroup$ – Parcly Taxel Mar 15 at 1:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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