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Thread: Weekly Practice Exam "P" Problems courtesy of Dr Krzysztof Ostaszewski

  1. #11
    Author, Instuctor and Seminar Provider krzysio's Avatar
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    Covariance

    Wat is exactly right. Or look at it this way: Covariance is a linear operator in each of its variables (This problem's purpose was to teach you that, comes in handy in all covariance problems on the test)

    Cov (X + Y, W) = Cov (X, W) + Cov (Y,W)
    Cov (X, Y+W) = Cov (X, Y) + Cov (X,Y)
    Cov (aX, W) =a Cov (X, W)
    Cov (X, aW) =a Cov (X, W)

    Does this help?

    Yours,
    Krzys'
    Last edited by krzysio; April 17th 2005 at 01:21 AM.
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  2. #12
    Author, Instuctor and Seminar Provider krzysio's Avatar
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    April 30 exercise

    Last edited by krzysio; May 19th 2009 at 11:01 PM.
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  3. #13
    Author, Instuctor and Seminar Provider krzysio's Avatar
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    Weibull distribution

    Yes, Weibull is on the test. And it is kind of hard to memorize all the distributions. So, to help you memorize the crucial item about the Weibull distribution, consider this exercise:

    The time to failure X of an MP3 player follows a Weibull distribution. It is known that and that the probability that this player stil works after 3 years is 1/e, while the same probability for 5 years is 1/(exp(4)). Find the probability that this MP3 player is still functional after 4 years.

    exp(4) means just e to the power 4.

    In other words, what is the survival function of the Weibull distribution, and can you memorize it?

    Yours,
    Krzys'
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  4. #14
    Author, Instuctor and Seminar Provider krzysio's Avatar
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    Exercise for April 30, 2005

    I have now posted this exercise with a solution:
    http://www.math.ilstu.edu/krzysio/4-...O-Exercise.pdf
    Enjoy.

    Yours,
    Krzys'
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  5. #15
    Author, Instuctor and Seminar Provider krzysio's Avatar
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    Exercise for May 7, 2005

    I have now posted this exercise with a solution:
    http://www.math.ilstu.edu/krzysio/5-7-5-KO-Exercise.pdf
    Enjoy.

    Yours,
    Krzys'
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  6. #16
    Author, Instuctor and Seminar Provider krzysio's Avatar
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    May 14 exercise for exam P

    I have posted a new exercise for exam P:
    http://www.math.ilstu.edu/krzysio/5-...O-Exercise.pdf
    This is in response to questions raised by some concerning transformations of random vectors, and sums of random variables, convolution and derivation of the PDF of the result of those operations. The exercise is given with three different solutions, showing three different approaches you can use for sums or differences of random variables: transformation of random vectors, convolution, or the CDF technique. It may feel a bit hairy, but it is meant to show you all of these techniques and help you deal with this kind of problems in every possible way. I hope you will enjoy.

    Yours,
    Krzys' Ostaszewski
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  7. #17
    Actuary.com - Level I Poster
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    April 30 problem

    We are asked to find the excess of Pr(X>11/4) over its bound.
    I don't understand why in the solution we apply the chebyshev's inequality to Pr((X-3/4|>30/16). And then find the prob of Pr(X>21/8), why do we need to calculate this probability?

    Shouldn't we rewrite the original Pr(X>11/4) to be
    Pr(|X-3/4|>8/4)=Pr(X>11/4)+Pr(X<-5/4)=Pr(X>11/4)

    Thanks.
    Last edited by jessie; May 1st 2005 at 10:13 PM.

  8. #18
    Author, Instuctor and Seminar Provider krzysio's Avatar
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    Chebyshev's Inequality

    Dear Jesse:

    You are absolutely right, I messed up. I posted a corrected version now:
    http://www.math.ilstu.edu/krzysio/4-...O-Exercise.pdf

    My apologies for this mess.

    Yours,
    Krzys'
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  9. #19
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    Variance/Covariance Matrix

    I have never seen a variance/covariance matrix before. Can you explain what the numbers in the matrix mean for the March 19, 2005 problem?

  10. #20
    Actuary.com - Level VI Poster
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    Quote Originally Posted by davenpoa
    I have never seen a variance/covariance matrix before. Can you explain what the numbers in the matrix mean for the March 19, 2005 problem?
    In a covariance matrix, the (i,j)th entry of the matrix = Cov(Xi,Xj).

    So, by properties of covariance and variance:

    1.) The matrix is symmetric (since Cov(Xi,Xj) = Cov(Xj,Xi)).
    2.) Cov(Xi,Xi) = Var(Xi).

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