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Conditional expectation of three exponential distributed r.v.

  1. Aug 27, 2010 #1
    I've been struggling with this problem for more than 4 days now:

    Let A, B and C be exponential distributed random variables with parameters lambda_A, lambda_B and lambda_C, respectively.

    Calculate E [ B | A < B < C ] in terms of the lambda's.

    I always seem get an integral which is impossible to calculate...

    Who has a suggestion how to solve this problem?
     
  2. jcsd
  3. Aug 28, 2010 #2
    That's how I'd try to do it too, can you show us what you got and where you were stuck?
     
  4. Aug 29, 2010 #3
    First a remark: we may assume that A, B and C are independent random variables (this may help a lot).

    [tex]
    E [ B | A < B < C ] = \int_{0}^\infty E [ B \, | \, A < B < C, A = a ] \; f_{A} ( a ) \; da
    [/tex]
    [tex]
    = \int_{0}^\infty \int_{0}^\infty E [ B \, | \, A < B < C, A = a, C=c ] \; f_{A} ( a ) \; f_{C} ( c ) \; da \; dc.
    [/tex]
    (Along with all that follows, this integral is only to be considered over R_+^2 and for c>a.)
    [tex]
    = \int_{0}^\infty \int_{0}^\infty E [ B \, | \, a < B < c, A = a, C=c ] \; f_{A} ( a ) \; f_{C} ( c ) \; da \; dc
    [/tex]
    (by independence)
    [tex]
    = \int_{0}^\infty \int_{0}^\infty E [ B \, | \, a < B < c ] \; f_{A} ( a ) \; f_{C} ( c ) \; da \; dc
    [/tex]

    We want to compute [tex] E [ B \, | \, a < B < c ]. [/tex]

    First note that
    [tex] E [ B \, | \, a < B < c ] = \frac{ \int_a^c t f_{B} (t) dt }{P( B \in [a,c])}.[/tex]

    We have[tex] \int_a^c t f_{B} (t) dt = ... = (a e^{- \lambda_B a} - c e^{- \lambda_B c}) + \frac{1}{\lambda_B} ( e^{- \lambda_B a} - e^{- \lambda_B c} ) [/tex]
    and
    [tex] P(B \in [a,c]) = F_B (c) - F_B (a) = ... = e^{- \lambda_B a} - e^{- \lambda_B c}. [/tex]

    Hence
    [tex] E [ B \, | \, a < B < c ] = \frac{ (a e^{- \lambda_B a} - c e^{- \lambda_B c}) + \frac{1}{\lambda_B} ( e^{- \lambda_B a} - e^{- \lambda_B c} ) }{e^{- \lambda_B a} - e^{- \lambda_B c}},[/tex]
    [tex] E [ B \, | \, a < B < c ] = \frac{ a e^{- \lambda_B a} - c e^{- \lambda_B c} }{e^{- \lambda_B a} - e^{- \lambda_B c}} + \frac{1}{\lambda_B}.[/tex]

    So
    [tex]
    E [ B | A < B < C ] = \int_{0}^\infty \int_{0}^\infty \left( \frac{ a e^{- \lambda_B a} - c e^{- \lambda_B c} }{e^{- \lambda_B a} - e^{- \lambda_B c}} + \frac{1}{\lambda_B} \right) \; f_{A} ( a ) \; f_{C} ( c ) \; da \; dc
    [/tex]
    [tex]
    = \int_{0}^\infty \int_{0}^\infty \left( \frac{ a e^{- \lambda_B a} - c e^{- \lambda_B c} }{e^{- \lambda_B a} - e^{- \lambda_B c}} \right) \lambda_A e^{- \lambda_A a} \; \lambda_C e^{- \lambda_C c} \; da \; dc +
    [/tex]
    [tex]
    \int_{0}^\infty \int_{0}^\infty \frac{1}{\lambda_B} \lambda_A e^{- \lambda_A a} \; \lambda_C e^{- \lambda_C c} \; da \; dc
    [/tex]


    Here's my problem: I can't compute the first of these integrals... And even then I'm not sure if what I'm doing here is correct... Is my calculation right, or is it wrong!? Is there perhaps a better way to calculate this? Etc.
     
    Last edited: Aug 29, 2010
  5. Aug 29, 2010 #4
    I'm not sure about this line - the marginals of the conditional distribution aren't necessarily exponential. Another approach is to apply directly the definition of conditional probabilities, so that

    [tex]E[B|A<B<C] = E[B.I(A<B<C)]/P[A<B<C][/tex]

    which is a ratio of integrals that should both be tractable.

    Hope this helps.
     
  6. Aug 29, 2010 #5
    Isn't this just the "law of total expectation"?

    I don't really understand your point I'm afraid.

    EDIT: Yes, we can also go by the second approach.

    Then how to calculate [tex] E[B.I(A<B<C)] [/tex] exactly ? (I'm not sure how to start with this integral.)
     
  7. Aug 29, 2010 #6
    I now understand what you mean by "I'm not sure about this line (...)": you're completely right, it's wrong!
     
  8. Aug 29, 2010 #7
    Ok, this can be simplified to

    [tex]\int_R b.f_A(a).f_B(b).f_C(c)da.db.dc[/tex]

    where R is the region a<b<c.
     
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