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Math Help - joint distribution

  1. #1
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    joint distribution

    Hi I have two problems on which I am stuck:

    In a vehicle servicing system, a vehicle arrives and is immediately placed in a service queue. It may be served immediately or it may have to wait in the queue. After waiting in the queue service, it is serviced and leaves immediately after being serviced. Let X be the tatal time, in minutes, the vehicle is in the system, waiting time plus service time. Let Y be the waiting time in the service queue. Assume X and Y are continuous random variable, where their bivariate space is{0<=y<=x and 0<=x}
    Assume the pdf of X and Y is:

    -f(x,y)=k*e^(x^2) , 0<=y<=x and 0<=x
    -0 elsewhere

    1- Find k.
    2- Find the conditional pdf of Y given X=x

    3- Find the expected value of Y given x=25.

    The problem is that f(x,y) does not seem to have y as variable. I found it difficult to work it .

    Thank you
    B
    PROBLEM 2

    Find the probability the gas station revenue is at least $6000( assume normality)

    please can you give me some hints?
    B
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by braddy View Post
    Hi I have two problems on which I am stuck:

    In a vehicle servicing system, a vehicle arrives and is immediately placed in a service queue. It may be served immediately or it may have to wait in the queue. After waiting in the queue service, it is serviced and leaves immediately after being serviced. Let X be the tatal time, in minutes, the vehicle is in the system, waiting time plus service time. Let Y be the waiting time in the service queue. Assume X and Y are continuous random variable, where their bivariate space is{0<=y<=x and 0<=x}
    Assume the pdf of X and Y is:

    -f(x,y)=k*e^(x^2) , 0<=y<=x and 0<=x
    -0 elsewhere

    1- Find k.
    The region over which f(x,y) is non-zero is (x>=0) and (0<=y<=x)

    then for f to be a pdf we require that:

    integral_{x=0 to infty} integral_{y=0 to x} f(x,y) dy dx = 1

    or:

    integral_{x=0 to infty} integral_{y=0 to x} k e^(x^2) dy dx = 1

    but this integral does not converge so presumably what was meant was:

    f(x,y) = k e^(-x^2),

    then

    integral_{x=0 to infty} integral_{y=0 to x} k e^(-x^2) dy dx = k/2,

    so as this must equal 1, we have k=2.

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by braddy View Post
    Hi I have two problems on which I am stuck:

    In a vehicle servicing system, a vehicle arrives and is immediately placed in a service queue. It may be served immediately or it may have to wait in the queue. After waiting in the queue service, it is serviced and leaves immediately after being serviced. Let X be the tatal time, in minutes, the vehicle is in the system, waiting time plus service time. Let Y be the waiting time in the service queue. Assume X and Y are continuous random variable, where their bivariate space is{0<=y<=x and 0<=x}
    Assume the pdf of X and Y is:

    -f(x,y)=k*e^(x^2) , 0<=y<=x and 0<=x
    -0 elsewhere

    1- Find k.
    2- Find the conditional pdf of Y given X=x
    If X=x then Y ~ U(0,x)

    3- Find the expected value of Y given x=25.
    and E(Y|X=25) = 12.5

    RonL
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  4. #4
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    Quote Originally Posted by braddy View Post
    PROBLEM 2

    Find the probability the gas station revenue is at least $6000( assume normality)

    please can you give me some hints?
    B
    There is not enough information to do this.

    RonL
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  5. #5
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    thank you CaptainBlack.

    But what do you mean by:
    If X=x then Y ~ U(0,x)
    ?
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  6. #6
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    Quote Originally Posted by braddy View Post
    thank you CaptainBlack.

    But what do you mean by:
    ?
    If the rv X takes value x then the conditional distribution of Y is the uniform
    distribution on (0,x).

    RonL
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  7. #7
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    CaptainBlack,

    Please can you be more explicit.
    for this answer
    Code:
     E(Y|X=25) = 12.5
    I tried but I cannot find it.

    Also for the same problem,
    How can I compute the marginal pdf of X

    I would do int{x=0 to x=infinty}2e^(-x^2)dx=1.77
    is it correct?
    B.
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  8. #8
    Grand Panjandrum
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    Let's hope I've done this right (needless to say this is not how I did it
    originaly - it help if you can spot what the answer has to be)

    E(Y|X=25) = integral_{y=0 to 25} y P(Y=y|X=25) dy

    but P(Y=y|X=25)P(X=25) = P(Y=y, X=25) = k e^{-25^2}

    but P(X=25) = integral_{y=0 to 25} k e^{-25^2} dy = k 25 e^{-25^2}

    so P(Y=y|X=25) = 1/25 for y in (0,25), and so:

    E(Y|X=25) = integral_{y=0,25} y/25 dy = (25^2/2)/25 = 12.5

    RonL
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by braddy View Post
    Also for the same problem,
    How can I compute the marginal pdf of X

    I would do int{x=0 to x=infinty}2e^(-x^2)dx=1.77
    is it correct?
    B.
    Marginal probabilty of X is:

    P(X=x) = integral_{y=0 to x} P(X=x,Y=y) dy = k x e*{-x^2}

    RonL
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