Are those solutions equivalent?

  • Thread starter fateswarm
  • Start date
  • #1
18
0

Main Question or Discussion Point

On my studying material I noticed that for "devices in parallel" problems, a question of the type "Find the probability the device works for 1 year" is dealt with P(A or B) where P(A), P(B) are the probabilities of each of the instruments working for one year, which is something I intuitively understand.

Now, when I went to a random variables chapter problem, for two random variables X and X, given their probability density function, for two devices in parallel where each device's life is represented by each random variable, the question "find the probability the device works for between 1 and 2 years" is dealt with the probability of max{X, Y} satisfying that criteria which appears to give a different result than the union of the corresponding probabilities!

I'll try to translate the actual problem: "The lifetimes X,Y (in hours) of two components are random variables with common probability density function fx,y(x,y) = e^(-x-y) with x,y>0. Calculate the probability the lifetime of the machine is between 1 and 2 hours when the components are connected in parallel." (it also asks for a connection in series). (It is also given from a previous exercise that the random variables are independent.)


My understanding is that if I use fx and fy and find the components probabilities separately and then calculate their union, the result would be the same. But in this case the result is: P(1<X<2)=P(1<Y<2)=0.2325 with P(1<X<2 or 1<Y<2) = 0.4109

While if I actually find the function for max{X,Y}, it's P(max{X,Y}<z) = P(X<z,Y<z) = P(X<z)P(Y<z) = Fx(x)Fy(y) which gives a different result.

Are the approaches equivalent?
 

Answers and Replies

  • #2
34,056
9,922
P(1<X<2)=P(1<Y<2)=0.2325 with P(1<X<2 or 1<Y<2) = 0.4109
That includes the case where X breaks between 1 and 2, but Y breaks afterwards (and vice versa) - so the device works longer than 2.
 
  • #3
18
0
That includes the case where X breaks between 1 and 2, but Y breaks afterwards (and vice versa) - so the device works longer than 2.
How? Aren't X, Y restricted to be examined between the times 1 and 2? How could X, or Y go beyond those time frames?
 
  • #4
18
0
I currently wonder if P(a<max{X,Y}<b) is equivalent to P(a<X<b or a<Y<b) to begin with. They seem to answer two different questions. The first being "the probability of the one with the maximum value being between those two values" and the latter "the probability of either being between those two values" effectively making the latter the right answer.

edit: Then again, I can't be certain since intuitively, either expression seems to cover it.

edit: The main indication is that I know P(A or B) was accepted in simpler problems.
 
Last edited:
  • #5
18
0
I think I figured it out, initially numerically and not intuitively, since I get the same result:

It appears that for the PARALLEL problem one has to test the MIN between X, and Y rather than the max. That way the result more conveniently is, for a simpler example of "less than 1 hour": P(min{X,Y}<1} = 1-P(min{X,Y}>1) =1-P(X>1,Y>1)=1-(1-P(X<1)(1-P(Y<1)) which is the same with the initial union of the simpler solutions.

If one goes with the MAX, the answer appears to be suitable for the intersection of the events and a problem of the components being in series. Now why would those problems work that way intuitively? I keep looking at it and there is little intuition that would e.g. point at the "min" of the two being suitable for the parallel problem, unless we're talking about an abstract concept of "I'm looking for at-a-minimum one of those two working in this interval" which is weird since it doesn't examine the interval itself but the existence of the random variables themselves in it, but I guess it's closer to reality.

edit: Or I guess, one I could say "I am satisfied with the minimum of those two being in this interval in order to answer the parallel problem", however "even the maximum of those two must satisfy the interval in order for them to work in series". Weird stuff when first seen but I guess it makes sense.
 
Last edited:
  • #6
18
0
The most intuitive way I found to look at it mathematically is to read the symbolism in reverse: e.g. P(y>max{X,Y}) for the serial problem. That way "y must be larger even than the larger of the two, hence both must be large enough [which is what I want in a serial situation]".
 

Related Threads on Are those solutions equivalent?

Replies
5
Views
766
  • Last Post
Replies
1
Views
7K
Replies
2
Views
2K
  • Last Post
Replies
0
Views
3K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
0
Views
2K
Replies
11
Views
896
Top