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Math Help - Exponential Distribution

  1. #1
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    Exponential Distribution

    The length of time Y necessary to complete a key operation in the construction has an exponential distribution with mean=10 hours. The formula C= 100 +40Y + 3Y^2 relates the cost C of completing this operation to the square of time to completion. Find the mean and variance of C.

    Don't know where to start here =(.
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    Quote Originally Posted by somestudent2 View Post
    The length of time Y necessary to complete a key operation in the construction has an exponential distribution with mean=10 hours. The formula C= 100 +40Y + 3Y^2 relates the cost C of completing this operation to the square of time to completion. Find the mean and variance of C.

    Don't know where to start here =(.
    Start by making the moment-generating function of this distribution your friend.

    Now note that C can be considered a linear function of 100, Y and Y^2. So:

    1. E(C) = 100 + 40 E(Y) + 3 E(Y^2).

    Get E(Y^2) from the moment generating function of Y - it's the second moment. Or just calculate it directly from the definition - in your case this reduces to finding 10 \int_{0}^{\infty} y^2 e^{-10y} \, dy (use repeated integration by parts).


    2. Var(C) = 3^2 Var(Y^2) + 40^2 Var(Y) + 2(3)(40) Cov(Y^2, Y).

    Note that Cov(Y^2, Y) = E(Y^3) - E(Y^2) E(Y).

    Get E(Y^3) from the moment generating function of Y - it's the third moment. Or just calculate it directly from the definition - in your case this reduces to finding 10 \int_{0}^{\infty} y^3 e^{-10y} \, dy (use repeated integration by parts).


    Details are left to you.


    Alternatively you could just work out the pdf for C and then use it to find E(C) directly from the definition. I'd suggest the method of transformations since the pdf for Y is decreasing in y.

    Then finding E(C^2) will enable you to calculate Var(C).
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    Thank you for the quick response. However i still feel pretty lost...

    I don't quite understand how you produced those integrals.
    When I solved the first integral i got 1, which is obviously not correct.

    Also Var(C) = 3^2 Var(Y^2) + 40^2 Var(Y) + 2(3)(40) Cov(Y^2, Y)
    what does cov stand for and why it is 3^2 40^2 and 2*3*40?

    Since moment generating function for exponential distribution is (1-Bt)^-1, it confuses me
    thanks again.
    Last edited by somestudent2; April 4th 2008 at 08:26 PM.
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    Quote Originally Posted by somestudent2 View Post
    Thank you for the quick response. However i still feel pretty lost...

    I don't quite understand how you produced those integrals.

    Also Var(C) = 3^2 Var(Y^2) + 40^2 Var(Y) + 2(3)(40) Cov(Y^2, Y)
    what does cov stand for and why it is 3^2 40^2 and 2*3*40?

    thanks again.
    1. Are you familiar with the concept of covariance?

    2. Are you familiar with the following definition?:

    E[g(Y)] = \int_{-\infty}^{\infty} g(y) \, f(y) \, dy , where g(Y) is a function of the random variable Y and f(y) is the pdf for the random variable Y.

    So, for example, E[Y^2] = \int_{-\infty}^{\infty} y^2 \, f(y) \, dy.
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    yes I am familiar with the second concept but not with covariance.
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    Quote Originally Posted by somestudent2 View Post
    yes I am familiar with the second concept but not with covariance.
    Well then, unless you get the pdf for the random variable C (using the technique I suggested, for example), I don't see how you can get the variance of C.
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    Hello again. I continued to work on this problem and looks like got through the first part. Got E(c)=104 3/50

    I have a question on how to find Var(Y^2) and Var(Y)? From theorem i know that Var(Y)=E(Y^2) - u^2, what would be then Var(Y^2) and our u here?

    would V(Y)=E(Y^2)-[E(Y)]^2

    and V(Y^2)=E(Y^3)-[E(Y^2)]^2?

    thanks
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    Quote Originally Posted by somestudent2 View Post
    Hello again. I continued to work on this problem and looks like got through the first part. Got E(c)=104 3/50

    I have a question on how to find Var(Y^2) and Var(Y)? From theorem i know that Var(Y)=E(Y^2) - u^2, what would be then Var(Y^2) and our u here?

    would V(Y)=E(Y^2)-[E(Y)]^2

    and V(Y^2)=E(Y^3)-[E(Y^2)]^2? Mr F says: V(Y^2)=E([Y^2]^2)-[E(Y^2)]^2 = E(Y^4)-[E(Y^2)]^2.

    thanks
    ..
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    ahhh i see, so I have to find the 4th moment as well, thanks
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