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Math Help - Clearification on Exponential Distribution

  1. #1
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    Clearification on Exponential Distribution

    Assume X_1 and X_2 are independent random variables, both following a exponential distribution. Calculate P\{X_1<X_2\}

    The way it's done in the book

    P\{X_1 < X_2\} = \int_0^\infty P\{X_1 < X_2|X_1 = x\} {\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx (a)

    = \int_0^\infty P\{ x< X_2\} {\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx (b)

    = \int_0^\infty e^{-\lambda_2x}{\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx (c)

    I'm wondering is why are they putting the pdf of exponential function (the part in blue)?

    in part (b) would is it \int_0^\infty \frac{P\{X_1 < X_2, X=x\}}{P\{X_1=x\}} {\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx \Longrightarrow \int_0^\infty \frac{P\{X_1 < X_2\} P\{X=x\}}{P\{X_1=x\}} {\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx \Longrightarrow \int_0^\infty P\{ x< X_2\} {\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx

    in part (c) isn't it P\{ x< X_2\} = \lambda_2 e^{-\lambda_2x}, therefore wouldn't it be: \int_0^\infty {\color [cmyk] {0.75,0,0.75,0.45}\lambda_2}e^{-\lambda_2x}{\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx

    why isn't there a {\color [cmyk] {0.75,0,0.75,0.45}\lambda_2}?
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  2. #2
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    Quote Originally Posted by lllll View Post
    Assume X_1 and X_2 are independent random variables, both following a exponential distribution. Calculate P\{X_1<X_2\}

    The way it's done in the book

    P\{X_1 < X_2\} = \int_0^\infty P\{X_1 < X_2|X_1 = x\} {\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx (a)

    = \int_0^\infty P\{ x< X_2\} {\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx (b)

    = \int_0^\infty e^{-\lambda_2x}{\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx (c)

    I'm wondering is why are they putting the pdf of exponential function (the part in blue)?

    in part (b) would is it \int_0^\infty \frac{P\{X_1 < X_2, X=x\}}{P\{X_1=x\}} {\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx \Longrightarrow \int_0^\infty \frac{P\{X_1 < X_2\} P\{X=x\}}{P\{X_1=x\}} {\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx \Longrightarrow \int_0^\infty P\{ x< X_2\} {\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx

    in part (c) isn't it P\{ x< X_2\} = \lambda_2 e^{-\lambda_2x}, therefore wouldn't it be: \int_0^\infty {\color [cmyk] {0.75,0,0.75,0.45}\lambda_2}e^{-\lambda_2x}{\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx

    why isn't there a {\color [cmyk] {0.75,0,0.75,0.45}\lambda_2}?
    Here is how I would do it:

    The joint pdf of X_1 and X_2 is f(x_1, x_2) = \lambda_1 \lambda_2 e^{-\lambda_1 x_1} e^{-\lambda_2 x_2}.

    Then \Pr(X_1 < X_2) = \int_{x_2 = 0}^{x_2 = +\infty} \int_{x_1 = 0}^{x_1 = x_2} f(x_1, x_2) \, dx_1 \, dx_2 = \, ....
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    just out of curiosity, if you were to add a third variable X_3 which is also independent and follows an exponential distribution, and wanted to calculate

    P\{X_1<X_2<X_3\} would you simply have a triple integral such as:

    P\{X_1<X_2<X_3\} = \int^{x_1=+\infty}_{x_1=0} \int^{x_2=+\infty}_{x_2=x_1} \int^{x_3=+\infty}_{x_3=x_2} f(x_1,x_2,x_3) \ dx_1 \ dx_2 \ dx_3

    where f(x_1,x_2,x_3) = \lambda_1 e^{-\lambda_1 x_1} \cdot \lambda_2 e^{-\lambda_2 x_2} \cdot \lambda_3 e^{-\lambda_3 x_3} ?
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  4. #4
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    Quote Originally Posted by lllll View Post
    just out of curiosity, if you were to add a third variable X_3 which is also independent and follows an exponential distribution, and wanted to calculate

    P\{X_1<X_2<X_3\} would you simply have a triple integral such as:

    P\{X_1<X_2<X_3\} = \int^{x_1=+\infty}_{x_1=0} \int^{x_2=+\infty}_{x_2=x_1} \int^{x_3=+\infty}_{x_3=x_2} f(x_1,x_2,x_3) \ dx_1 \ dx_2 \ dx_3

    where f(x_1,x_2,x_3) = \lambda_1 e^{-\lambda_1 x_1} \cdot \lambda_2 e^{-\lambda_2 x_2} \cdot \lambda_3 e^{-\lambda_3 x_3} ?
    I'd calculate it as \Pr(X_2 < X_3) - \Pr(X_2 < X_1).
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    Quote Originally Posted by mr fantastic View Post
    I'd calculate it as \Pr(X_2 < X_3) - \Pr(X_2 < X_1).
    Not quite, since this even can be negative... It should be:
    P(X_1<X_2<X_3)=P(X_2<X_3)-P(X_2<X_3\mbox{ and }X_2\leq X_1) =P(X_2<X_3)-P(X_2<\min(X_1,X_3))
    and it is well known that \min(X_1,X_3) is an exponential random variable of parameter \lambda_1+\lambda_3, and it is independent of X_2 so, in application of the result of your initial question,
    P(X_1<X_2<X_3)=\frac{\lambda_2}{\lambda_2+\lambda_  3}-\frac{\lambda_2}{\lambda_1+\lambda_2+\lambda_3}=\f  rac{\lambda_1\lambda_2}{(\lambda_1+\lambda_2)(\lam  bda_1+\lambda_2+\lambda_3)}

    One easily checks that the integral gives the same result. By the way, the computation of the integral gives by induction the general formula:
    P(X_1<\cdots<X_n)=\frac{\lambda_1\cdots\lambda_{n-1}}{(\lambda_n+\lambda_{n-1})(\lambda_n+\lambda_{n-1}+\lambda_{n-2})\cdots(\lambda_n+\cdots+\lambda_1)}
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  6. #6
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    Quote Originally Posted by Laurent View Post
    Not quite, since this even can be negative... It should be:
    P(X_1<X_2<X_3)=P(X_2<X_3)-P(X_2<X_3\mbox{ and }X_2\leq X_1) =P(X_2<X_3)-P(X_2<\min(X_1,X_3))
    and it is well known that \min(X_1,X_3) is an exponential random variable of parameter \lambda_1+\lambda_3, and it is independent of X_2 so, in application of the result of your initial question,
    P(X_1<X_2<X_3)=\frac{\lambda_2}{\lambda_2+\lambda_  3}-\frac{\lambda_2}{\lambda_1+\lambda_2+\lambda_3}=\f  rac{\lambda_1\lambda_2}{(\lambda_1+\lambda_2)(\lam  bda_1+\lambda_2+\lambda_3)}

    One easily checks that the integral gives the same result. By the way, the computation of the integral gives by induction the general formula:
    P(X_1<\cdots<X_n)=\frac{\lambda_1\cdots\lambda_{n-1}}{(\lambda_n+\lambda_{n-1})(\lambda_n+\lambda_{n-1}+\lambda_{n-2})\cdots(\lambda_n+\cdots+\lambda_1)}
    Ha. Thanks for the catch. Obvious in hindsight, I'm annoyed I missed it the first time.
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