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  1. #1
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    exponential distribution question

    I'm having trouble setting up the following equation:
    The length of time Y necessary to complete a key operation in the construction of houses has an exponential distribution with mean 10 hours. The formula C=100+4Y+3Y^2 relates to the cost of C. Find the mean of C.

    initially would it be:

    \frac{1}{\beta}e^{-\frac{y}{\beta}} = \frac{1}{10}e^{-\frac{y}{10}}

    \int^{\infty}_0 y \frac{1}{10}e^{-\frac{y}{10}} \times (100+4y+3y^2) \ dy

    then just integrate out?
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    Quote Originally Posted by lllll View Post
    I'm having trouble setting up the following equation:
    The length of time Y necessary to complete a key operation in the construction of houses has an exponential distribution with mean 10 hours. The formula C=100+4Y+3Y^2 relates to the cost of C. Find the mean of C.

    initially would it be:

    \frac{1}{\beta}e^{-\frac{y}{\beta}} = \frac{1}{10}e^{-\frac{y}{10}}

    \int^{\infty}_0 y \frac{1}{10}e^{-\frac{y}{10}} \times (100+4y+3y^2) \ dy

    then just integrate out?
    No:

    1. It's wrong. It should be \int^{\infty}_0 \frac{1}{10}e^{-\frac{y}{10}} \times (100+4y+3y^2) \ dy (note the missing y at the front).

    2. There's an easier way:

    E[C] = E[100+4Y+3Y^2] = E[100] + 4 E[Y] + 3 E[Y^2] = 100 + 4 (10) + E[Y^2].

    Now recall that Var[Y] = E[Y^2] - (E[Y])^2. And you should know that for an exponential distribution, Var[Y] = (E[Y])^2. Therefore E[Y^2] = 2 (E[Y])^2 = 200.

    If you want you can also get E[Y^2] by integration.
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    Quote Originally Posted by mr fantastic View Post
    No:

    1. It's wrong. It should be \int^{\infty}_0 \frac{1}{10}e^{-\frac{y}{10}} \times (100+4y+3y^2) \ dy (note the missing y at the front).

    2. There's an easier way:

    E[C] = E[100+4Y+3Y^2] = E[100] + 4 E[Y] + 3 E[Y^2] = 100 + 4 (10) + E[Y^2].

    Now recall that Var[Y] = E[Y^2] - (E[Y])^2. And you should know that for an exponential distribution, Var[Y] = (E[Y])^2. Therefore E[Y^2] = {\color{red}2} (E[Y])^2 = 200.

    If you want you can also get E[Y^2] by integration.
    Where did the 2 come from?
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    Nevermind, I figured it out. But on a side note would finding the Variance of the cost follow a similar approach?
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    Quote Originally Posted by lllll View Post
    Nevermind, I figured it out. But on a side note would finding the Variance of the cost follow a similar approach?
    Yes, but it'll be more tedious and there's no avoiding some integration:

    Var[C] = E[C^2] - (E[C])^2.

    You already know E[C] so you have half the answer. The easy half, unfortunately.

    C^2 = (100 + 4Y + 3Y^2)^2 = \, ........

    To get E[Y^3] and E[Y^4] you'll have to integrate. (Alternatively you could use the moment generating function).
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    if you were to integrate it out would it be:

    \int^{\infty}_0 y^2\frac{1}{10}e^{-\frac{y}{10}} \times (100+4y+3y^2)^2 \ dy

    =\int^{\infty}_0 y^2\frac{1}{10}e^{-\frac{y}{10}} \times (10,000+8,000y+660y^2+24y^3+9y^4) \ dy

    =\int^{\infty}_0 y^2\frac{1}{10}e^{-\frac{y}{10}} \times (10,000+8,000(10)+660(220)+24y^3+9y^4) \ dy ?
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  7. #7
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    Quote Originally Posted by lllll View Post
    if you were to integrate it out would it be:

    \int^{\infty}_0 y^2\frac{1}{10}e^{-\frac{y}{10}} \times (100+4y+3y^2)^2 \ dy

    =\int^{\infty}_0 y^2\frac{1}{10}e^{-\frac{y}{10}} \times (10,000+8,000y+660y^2+24y^3+9y^4) \ dy

    =\int^{\infty}_0 y^2\frac{1}{10}e^{-\frac{y}{10}} \times (10,000+8,000(10)+660(220)+24y^3+9y^4) \ dy ?
    No. E \left[C^2\right] = \int^{\infty}_0 \frac{1}{10}e^{-\frac{y}{10}} \times (100+4y+3y^2)^2 \ dy. This time you have a y^2 at the front that shouldn't be there.

    But why do all this when all you really need to do is use integration to find these two things:

    E[Y^3] = \int^{\infty}_0 y^3 \, \frac{1}{10}e^{-\frac{y}{10}} ~ dy

    E[Y^4] = \int^{\infty}_0 y^4 \, \frac{1}{10}e^{-\frac{y}{10}} ~ dy
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