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Math Help - Convolution of Exponential Random Variables

  1. #1
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    Convolution of Exponential Random Variables

    Let  T_1, T_2 be independent exponential variables with parameters  \lambda_1, \lambda_2 respectively. Find the density function of  R = T_1 - T_2.

    I get a piecewise density function (for r > 0, r \leq 0):

     <br />
 f_R (r) = \frac{\lambda_1\lambda_2}{\lambda_1+\lambda_2} e^{{-\lambda_2}r}, r>0

    f_R (r) = \frac{\lambda_1\lambda_2}{\lambda_1+\lambda_2} e^{{\lambda_1}r},  r\leq 0

    The answer given however is simply that the density function is exponential without further information. I was wondering if my answer is correct?

    P.S. I can't seem to do piecewise using normal LaTeX code; is it different in these forums?
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  2. #2
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    The first one should have e^{-\lambda_1 r} I think but other than that it seems fine.

    If you have problems with latex, post it in the latex subforum, 'tis working fine for me.
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  3. #3
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    Quote Originally Posted by h2osprey View Post
    Let  T_1, T_2 be independent exponential variables with parameters  \lambda_1, \lambda_2 respectively. Find the density function of  R = T_1 - T_2.

    I get a piecewise density function (for r > 0, r \leq 0):

     <br />
 f_R (r) = \left\{\begin{array}{c}\frac{\lambda_1\lambda_2}{\  lambda_1+\lambda_2} e^{{-\lambda_2}r}, r>0\\ \frac{\lambda_1\lambda_2}{\lambda_1+\lambda_2} e^{{\lambda_1}r},  r\leq 0\end{array}\right.

    The answer given however is simply that the density function is exponential without further information. I was wondering if my answer is correct?

    P.S. I can't seem to do piecewise using normal LaTeX code; is it different in these forums?
    There is a mistake in the exponents: it is e^{-\lambda_1 r} when r>0 and e^{-\lambda_2 r} else.

    How to find the result with little computation (but some nontrivial knowledge): first, the probability that T_1>T_2 is \frac{\lambda_2}{\lambda_1+\lambda_2} (classic useful computation); then, by the memoryless property of exponential distribution, conditionally to T_2 and to the event \{T_1>T_2\}, the distribution of T_1-T_2 is exponential with parameter \lambda_1 (note that since this distribution does not depend on T_2, it was unnecessary to condition by T_2). This gives the " r>0" part: \frac{\lambda_2}{\lambda_1+\lambda_2}\lambda_1 e^{-\lambda_1 r}, and symmetry gives the other part.
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    Quote Originally Posted by Laurent View Post
    There is a mistake in the exponents: it is e^{-\lambda_1 r} when r>0 and e^{-\lambda_2 r} else.

    How to find the result with little computation (but some nontrivial knowledge): first, the probability that T_1>T_2 is \frac{\lambda_2}{\lambda_1+\lambda_2} (classic useful computation); then, by the memoryless property of exponential distribution, conditionally to T_2 and to the event \{T_1>T_2\}, the distribution of T_1-T_2 is exponential with parameter \lambda_1 (note that since this distribution does not depend on T_2, it was unnecessary to condition by T_2). This gives the " r>0" part: \frac{\lambda_2}{\lambda_1+\lambda_2}\lambda_1 e^{-\lambda_1 r}, and symmetry gives the other part.
    Yes, turns out I switched the numbers while typing the problem: the original question was R = T_2 - T_1 . However, regarding the pdf when r \leq 0 shouldn't there be a negative sign? If not it gets very, very large indeed..

    and thanks for the LaTeX code!
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  5. #5
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    Quote Originally Posted by h2osprey View Post
    However, regarding the pdf when r \leq 0 shouldn't there be a negative sign? If not it gets very, very large indeed..
    Yes, sure, it's e^{\lambda_2 r} for r<0; the only mistake was with indices 1 and 2, due to switch them in the definition.
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