0
$\begingroup$

Let $X$ be the lifetime of a personal computer. Suppose the pdf of $X$ is $$ f_X(x) = \begin{cases}\frac{1}{5}e^{-\frac{1}{5}x} & x > 0 \\ 0 & {\rm otherwise}\end{cases} $$ Assume that costumers replace their computers at failure or three years, whichever comes first. Let the actual life of a computer, which denoted the duration of a computer kept by an costumer before replacement.

(a) Find the cdf of $F_T(t)$ of the actual life $T$ of a computer and then write it in one formula using the unit step functions

I learn that i can get the cdf by integrating the pdf. $$\int 1/5e^{-t/5} \,dt= -e^{-t/5}$$ my way to get the cdf : \begin{cases} 0 & \text{$t<0$} \\ -e^{-t/5} & \text{0≤t≤3}\\ 1 & \text{$t>3$} \end{cases}$ solution: \begin{cases} 0 & \text{$t<0$} \\ 1-e^{-t/5} & \text{0≤t≤3}\\ 1 & \text{$t>3$} \end{cases}$

Hope someone can explain why it is $1-e^{-1/5t}$ and the meaning of it.

$\endgroup$
3
  • $\begingroup$ Please phrase your question properly. $\endgroup$
    – Sonal_sqrt
    Commented Feb 21, 2018 at 6:55
  • $\begingroup$ what is "the lifetime of the computer for 2 years"? $\endgroup$
    – Cettt
    Commented Feb 21, 2018 at 7:05
  • $\begingroup$ Please, it is $e^{-t/5}$ not $e^{-1/5t}$. $\endgroup$ Commented Feb 21, 2018 at 7:27

2 Answers 2

0
$\begingroup$

Note that

$t < 0$

$$ \int_{-\infty}^{t} {\rm d}t~ f_T(t) = 0 $$

$t>0$

$$ \int_{-\infty}^{t} {\rm d}t~ f_T(t) = \int_{-\infty}^{0} {\rm d}t~ f_T(t) + \int_{0}^{t} {\rm d}t~ f_T(t) = 0 + \int_{0}^{t} {\rm d}t~ \frac{1}{5}e^{-t/5} = 1 - e^{-t/5} $$

So in summary

$$ F_T(t) = \left\{\begin{array}{ccc} 0 &,& t < 0 \\ 1 - e^{-t/5} &,& t\ge 0\end{array}\right. $$

$\endgroup$
0
$\begingroup$

You should integrate from $0$ to $t$. This gives you $$ \frac{1}{5} \int_0^t e^{-\frac{1}{5}x} = 1-e^{-\frac{1}{5}t}. $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .