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My question is related to my job and asking to learn.

Simplified case is:

Event A and Event B failure rates are exponentially distributed.

Event A (failure of item A) => failure rate (lambda): 0,0002

Event B (failure of item A) => failure rate (lambda): 0,0008

for a duration of 500.

I want to understand the probability that both events occurs but event A occurs before event B.

Event A and B are independent. One does not trigger (not change the prob.) the other.

I made 200,000 trials monte carlo sim. and get 0,01533 for probability that both events occurs but event A occurs before event B.

Q1) But where is this coming from? What is the analytical solution for this case?

Q2) Which specific statistics topic should I read on web if the number of sequential items increases and the problem gets more complex?

thanks

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  • $\begingroup$ Do you know how to apply conditioning? $\endgroup$ – soakley Feb 28 '17 at 18:17
  • $\begingroup$ Hi Soakley, thanks for dealing. If you mean conditional prob., yes I'm aware of it but here I think Prob B given that A already occured which is P(B) because of independence won't help me. I can understand the answer (if provided) but I'm stucked to create mine since I'm an engineer (not a statistician) and it's been 15-16 years since I got 2 semester of statistics courses. Can you please lead me? Thanks again. $\endgroup$ – Andre Chenier Mar 1 '17 at 4:53
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If we let $X$ be the time until event A and $Y$ be the time until event B, then both $X$ and $Y$ have exponential distributions (let's call their rates $\lambda_1$ and $\lambda_2,$ respectively) and it appears you want to find $P[X < Y, X<500,Y<500]$

We know by independence that the probability that both $X$ and $Y$ are less than 500 is given by $T=\left( 1-e^{-500 \lambda_1} \right) \left( 1-e^{-500 \lambda_2} \right)$

I'm not sure why I did it this way, but within the domain (a square) of both variables being less than 500 I found the complementary area $U=P[X>Y, X<500, Y<500]$ to what you are interested in by using a double integration.

Then your desired quantity is $T-U.$

I found that to be

$$T-U =\frac{1}{\lambda_1+\lambda_2} \left[ \lambda_1 + \lambda_2 e^{-500 \left( \lambda_1 + \lambda_2 \right) } \right]-e^{-500 \lambda_2} $$

For your rates this evaluates to $0.0149045,$ in good agreement with your simulation result.

Now let's set up the direct approach. We want to evaluate the probability that $X<Y$ constrained by both variables being no more than 500. Call the answer $Q.$

$$Q=\int_0^{500}\int_x^{500}\lambda_1 e^{-\lambda_1x} \lambda_2e^{-\lambda_2y} \ dy \ dx$$

If you solve this, you will get the same answer as above. Can you take it from here?

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  • $\begingroup$ I think the event of interest is not $X<Y$ but $X<Y<500$ $\endgroup$ – Juho Kokkala Mar 1 '17 at 18:35
  • $\begingroup$ Ah, I couldn't figure what the "duration of 500" meant. $\endgroup$ – soakley Mar 1 '17 at 18:41
  • $\begingroup$ @soakley It's good to see that my simulation result is acceptable and the solution has a closed form. However what I couldn't understand was how to deal with P(X<Y) and I didn't understand the U. I suppose integrals are using the PDF function so I also couldn't get the idea of double integral because in my mind a double integral of a line will give a volume and I couldn't match a volume with my question logically. (I drawed the PDFs and tried to see the geometry but...) Sorry for bothering and taking your time. $\endgroup$ – Andre Chenier Mar 2 '17 at 5:30
  • $\begingroup$ @soakley Can you please give the integrals or describe the geometry in your mind for U or for P(X<Y). I can get it in that way. As I've said I want to understand the logic, not only got the answer in numbers. Thanks for your efforts. $\endgroup$ – Andre Chenier Mar 2 '17 at 5:31

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