0
$\begingroup$

$A$: random variable to denote time to failure for equipment E_A

$B$: random variable to denote time to failure for equipment E_B

$A$ and $B$ are independent.

$P(A◁B)(T)$ : Probability that A realizes before B in a mission whose duration is T unit.

example scenario:

$A$ and $B$ both exhibit exponential distribution.

E_A failure rate: $\lambda_A=5E-3$ /hours

E_B failure rate: $\lambda_B=9.4E-3$ /hours

duration: $100$ hours

what is $P(A◁B)(T)$ ?

I tried: (found equation below in a paper. I can understand the equation.)

$$P(A◁B)(T)=\int_0^T \biggl( pdf_A(t)*(1-CDF_B(t)) \biggr) dt$$

my questions:

  1. integral result is 0.2650 if $\int$ time range is $0$ to $100$
  2. integral result is 0.3472 if $\int$ time range is $0$ to $\infty$
  3. Which one is correct, $0.2650$ or $0.3472$? Why?

my confusion to ask question 3 is why result depends on time? I feel that only the values of $\lambda$s are significant here since exponential rate is constant and exponential distr. is memoryless.

it's been >15 years since my statistics lessons. May you please help? regards

$\endgroup$
3
  • $\begingroup$ What if time is 1 hour? Or half an hour? Then you likely end up with a situation where neither fail in which case A does not fail before B. This is why you need to time into consideration. $\endgroup$
    – num_39
    Commented Feb 18, 2022 at 7:31
  • $\begingroup$ Even for time = 100, there's the possibility that neither fail in which case A does not fail before B. So you need to take that possibility into account as well. $\endgroup$
    – num_39
    Commented Feb 18, 2022 at 7:33
  • $\begingroup$ sure. 1st mission (0 to100)=26.5% , 2nd mission: 6.16% , 3rd mission 0.0616 , 4th mission 0.0146 and so on. Sum of these probabilities is 34.72%. In my real life related question I don't have any $T_0$ or $T_1$. I just have duration $\endgroup$
    – Ersin
    Commented Feb 18, 2022 at 7:53

1 Answer 1

1
$\begingroup$

The reason your calculation in 1 is smaller than your calculation in 2 is that it is a more specific event. The respective events for these probability calculations are:

$$\begin{matrix} 1. & & & & & A < B \ \text{ and } A < T \\[6pt] 2. & & & & & A < B. \\[6pt] \end{matrix}$$

As to which is correct, that depends on which event you are interested in.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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