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Conditional expectation of exponential distribution.

  1. Mar 9, 2010 #1
    I have been stuck at this calculation. There are two exponential distributions X and Y with mean 6 and 3 respectively. We need to find
    I keep getting it negative, which is clearly wrong. Anybody wants to try it?
  2. jcsd
  3. Mar 10, 2010 #2


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    Show your calculation. It should be easy to find the flaw.
  4. Mar 10, 2010 #3
    I got it finally. But here is a brief note about it

    [tex]E[t_2-t_1|t_2>t_1] = E[(t_2-t_1)*I_{\{t_2>t_1\}}] = \int_0^{\infty} \int_{t_1}^{\infty}(t_2-t_1)\lambda_1 \lambda_2 e^{-\lambda_1 t_1} e^{-\lambda_2 t_2} dt_2 dt_1[/tex]

    This nicely gives the answer as [tex]\frac{\lambda_1}{\lambda_2(\lambda_1+\lambda_2)}[/tex]

    I was doing a mistake in byparts.
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