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Joint density problem.

  1. Feb 7, 2012 #1
    1. The problem statement, all variables and given/known data

    I have:

    [tex]f_A=\lambda e^{-\lambda a}[/tex]
    [tex]f_B=\mu e^{-\mu b}[/tex]

    ([itex]A[/itex] and [itex]B[/itex] are independent)

    I need to find the density of [itex]C=\min(A,B)[/itex]

    2. The attempt at a solution
    [tex]f_C(c)=f_A(c)+f_B(c)-f_A(c)F_B(c)-F_B(c)f_A(c)[/tex]
    [tex]=\lambda e^{-\lambda c}+\mu e^{-\mu c}-\lambda e^{-\lambda c}(1-e^{-\mu c})-(1-e^{-\lambda c})\mu e^{-\mu c}[/tex]
    [tex]=\lambda e^{-\lambda c}e^{-\mu c}+\mu e^{-\lambda c}e^{-\mu c}[/tex]
    [tex]=2(\lambda+\mu)e^{-c(\lambda+\mu)}[/tex]

    Correct or utterly wrong?
     
  2. jcsd
  3. Feb 7, 2012 #2

    Ray Vickson

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    Homework Helper

    Some blunders (probably just typos), but answer is right: your first line should have been
    [tex]f_C(c)=f_A(c)+f_B(c)-f_A(c)F_B(c)-F_A(c)f_B(c),[/tex]
    which is what your later lines computed. However, you are doing it the hard way: much easier is to say [tex] \Pr \{\min(A,B) > c \} = \Pr \{ A > c \mbox{ and } B > c \}
    = \Pr \{A > c \} \cdot \Pr \{ B > c \}. [/tex]

    RGV
     
  4. Feb 8, 2012 #3
    [tex]=(1-F_A(c)) \cdot (1-F_B(c))=(1-(1-e^{-\lambda c})) \cdot (1-(1-e^{-\mu c}))=e^{-c(\lambda+\mu)}[/tex]
    [tex]\Rightarrow\frac{d}{dc}e^{-c(\lambda+\mu)}=-(\lambda+\mu)e^{-c(\lambda+\mu)}[/tex]

    Although my first answer should have been: [tex](\lambda+\mu)e^{-c(\lambda+\mu)}[/tex]

    Which is correct?
     
    Last edited: Feb 8, 2012
  5. Feb 8, 2012 #4

    Ray Vickson

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    Science Advisor
    Homework Helper

    Your first *answer* was incorrect, but your method was correct up to the second-last line; in my original response, I messed the factor of 2, so should not have said it was correct. Basically, to get your last line you said a+b = 2(a+b), so you made a blunder.

    The result [itex] \min(A,B) \leftrightarrow (\lambda+\mu) e^{-(\lambda + \mu)t} [/itex] is correct. It is one of the absolutely standard properties of the exponential. Since the second way of getting it is correct, step-by-step, it cannot fail to be correct; you just need more confidence when making true statements, but you also need to be careful when doing algebraic manipulations.

    RGV
     
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