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Exponential Random Variables and Conditional Probability Problem

  1. Mar 27, 2012 #1
    1. The problem statement, all variables and given/known data
    A submarine has three navigational devices but can remain at sea if at least two are working. Suppose that the failure times are exponential with means 1 year, 1.5 years, and 3 years. What is the average length of time the boat can remain at sea?


    2. Relevant equations
    Density for an exponential random variable: f(x)= λe-λx for x>=0
    E(T)=1/λ if T is an exponential random variable

    Maybe relevant: P(S<T)=λS / (λS + λT) for exponential random variables S and T, and this is similar for many exponential random variables.

    3. The attempt at a solution

    The boat can remain at sea until 2 parts break.
    Let T be the time that the boat can remain at sea.

    Since the sample space can be divided into the order in which the parts fail, we have:

    E(T)= Sum where 1<=i,j,k<=3 of: E(T|Ti<Tj<Tk)P(Ti<Tj<Tk)

    where E(T|Ti<Tj<Tk)=E(Tj|Ti<Tj<Tk) since the boat can remain at sea until two parts fail.

    Now, this would be a similar method that we've used with discrete random variables in class. Unfortunately, I'm a little rusty with my continuous probability. Is there somewhere to go from here, or maybe an easier way to approach the problem?

    I very much appreciate any help you can give!

    Thanks!
     
  2. jcsd
  3. Mar 27, 2012 #2
    We ask ourselves: What is the probability the boat lasts less than or equal to t years? If we have an expression for this probability, we can find the expectation as the weighted sum of all years by the probability for that year (in a continuum through integration).

    [tex]P(T \le t) = P((T_1 \le t \cap T_2 \le t) \cup (T_1 \le t \cap T_3 \le t) \cup (T_2 \le t \cap T_3 \le t)) [/tex]

    Using set theory. Let A B and C be events:
    [tex]P(A \cup B \cup C)=P(A) + P(B)+P(C)-P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)[/tex]

    This results in, I believe, if T_1, T_2, and T_3 are independent variables:
    [tex] E\{T\} = e_{t1}+e_{t2}+e_{t3}-e_{t1}e_{t2}-e_{t1}e_{t3}-e_{t2}e_{t3}+e_{t1}e_{t2}e_{t3}[/tex]
    where
    [tex] E\{ T_j \} = e_{tj}[/tex]

    edit: Upon further thought, I don't think I am right. GL.
     
    Last edited: Mar 27, 2012
  4. Mar 27, 2012 #3
    Yea, in that case, the E[T]=1 when the answer in the back of the book is E[T]=1.35 years

    Thanks for trying!
     
  5. Mar 28, 2012 #4

    Ray Vickson

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    Science Advisor
    Homework Helper

    The problem is quite easy if you know the memoryless and decomposition properties of exponential processes, and harder if you don't. Here is what I mean: let the three independent failure times be exponential with rates a, b and c. The rates have u nits of "per year" and are a = 1. b = 2/3 and c = 1/3 in your case; they are the reciprocals of the means. For simplicity of writing, denote the exponential density u*exp(-u*t) as fe(t,u).

    Anyway, the time to first failure is the minimum of the three, so is exponential with rate = sum of all the rates; that is, T1~ has density f1(t) = fe(t,a+b+c). The time to the next failure depends on which of the three units failed first. By the "decomposition property", the probability that unit A failed first is a/s, where s = a+b+c. Given that A failed first, the times to failure of B and C are still exponential with their original parameters; that is, B and C are "like new"---that is the memoryless property. So, the time to the next failure is the min of (Tb,Tc), so has density fe(t,b+c). Similarly if B or C fail first, etc. Thus, the time to the next failure is a mixture, with density f2(t) = (a/s)*fe(t,b+c) + (b/s)*fe(t,a+c) + (c/s)*fe(t,a+b).
    The time to failure of the sub has density fs(t) which is the convolution of f1 and f2. However, all you really want is the expected value, which is a bit simpler.

    Anyway, you now have all the machinery you need to get the answer; doing convolutions of exponentials is easy. When you work it all out you will, indeed, obtain the answer that the mean time to failure of the sub is 1.35 years.

    RGV
     
  6. Mar 29, 2012 #5
    Thanks Ray! You're very good at explaining things, and I appreciate your help (again).
     
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