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Math Help - Quick question on exponential distribution

  1. #1
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    Quick question on exponential distribution

    I just wanted to make sure I was doing these problems correctly, as I don't have the solutions to any of them, or previous work to base my answers on.

    given an exponential distribution compute:

    a) E[X^2 -2X|X=1]

    what I was thinking it's simply \int_1^\infty \lambda e^{-\lambda X} \cdot X^2 \ dX - 2\int_1^\infty \lambda e^{-\lambda X} \cdot X \ dX


    b) P\{X>1|X<3\} = \int_1^{\infty} \lambda e^{-\lambda x} \ dx \cdot \bigg{(} \int_0^3 \lambda e^{-\lambda x} \ dx \bigg{)}^{-1}


    c) P\{X_3>X_1+X_2\} = \int_{X_1=0}^{X_1=\infty} \int_{X_2=0}^{X_2=\infty}\int_{X_3=X_1+X_2}^{X_3=\  infty}\lambda_1 e^{-\lambda_1 X_1} \lambda_2 e^{-\lambda_2 X_2}  \lambda_3 e^{-\lambda_3 X_3} \ dX_3 \ dX_2 \ dX_1


    d) compute P\{X_1=\min\{X_1, \ X_2\}\} with rates \lambda_1 and \lambda_2

    based on my book the probabilistic minimum of is just \frac{1}{\sum_{i=1}^n \lambda_i} (no proof was given) so in this case it would be \frac{1}{\lambda_1+\lambda_2} also I would like to show that \frac{1}{\sum_{i=1}^n \lambda_i} is the case, with 2 variable. I thought it would be:

    P\{X_1, < X_2\} + P\{X_2, < X_1\} but that give me 1 and

    another way I considered was P\{X_1, < X_2\} \cdot P\{X_2, < X_1\} = \frac{\lambda_1 \lambda_2}{(\lambda_1 +\lambda_2)^2}

    I finally considered n[1-F(y)]^{n-1} \cdot f(y) but I don't know what \lambda to pick since they are both different.
    Last edited by lllll; December 14th 2008 at 10:28 PM.
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  2. #2
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    Quote Originally Posted by lllll View Post
    I just wanted to make sure I was doing these problems correctly, as I don't have the solutions to any of them, or previous work to base my answers on.

    given an exponential distribution compute:

    a) E[X^2 -2X|X=1]

    what I was thinking it's simply \int_1^\infty \lambda e^{-\lambda X} \cdot X^2 \ dX - 2\int_1^\infty \lambda e^{-\lambda X} \cdot X \ dX
    Did you write the question correctly? Given your answer, it looks likes you were asked to compute E[X^2-2X|X\geq 1]. Anyway, your answer equals E[(X^2-2X){\bf 1}_{(X\geq 1)}], i.e. you forgot to divide the result by P(X\geq 1).

    (By the way, it is usually advisable to write small x in the integral because it is not exactly the same as X, without further precision)

    b) P\{X>1|X<3\} = \int_1^{\infty} \lambda e^{-\lambda x} \ dx \cdot \bigg{(} \int_0^3 \lambda e^{-\lambda x} \ dx \bigg{)}^{-1}
    Always start by writing what these conditional expectations stand for; in this case, it is P(X>1|X<3)=\frac{P(X>1\mbox{ and }X<3)}{P(X<3)}. Can you see why I said that?


    c) P\{X_3>X_1+X_2\} = \int_{X_1=0}^{X_1=\infty} \int_{X_2=0}^{X_2=\infty}\int_{X_3=X_1+X_2}^{X_3=\  infty}\lambda_1 e^{-\lambda_1 X_1} \lambda_2 e^{-\lambda_2 X_2}  \lambda_3 e^{-\lambda_3 X_3} \ dX_3 \ dX_2 \ dX_1
    OK, for a start.

    d) compute P\{X_1=\min\{X_1, \ X_2\}\} with rates \lambda_1 and \lambda_2

    based on my book the probabilistic minimum of is just \frac{1}{\sum_{i=1}^n \lambda_i} (no proof was given) so in this case it would be \frac{1}{\lambda_1+\lambda_2} also I would like to show that \frac{1}{\sum_{i=1}^n \lambda_i} is the case, with 2 variable. I thought it would be:

    P\{X_1, < X_2\} + P\{X_2, < X_1\} but that give me 1 and

    another way I considered was P\{X_1, < X_2\} \cdot P\{X_2, < X_1\} = \frac{\lambda_1 \lambda_2}{(\lambda_1 +\lambda_2)^2}

    I finally considered n[1-F(y)]^{n-1} \cdot f(y) but I don't know what \lambda to pick since they are both different.
    I don't get it; you want P\{X_1=\min\{X_1, \ X_2\}\}. Isn't it just the same as P\{X_1\leq X_2\}?
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