Thank you Dr. Krzysztof Ostaszewski for suggesting we post some of your sample problems. . Watch for further exam problems as he makes them available to us.
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Weekly Practice Exam "P" Problems courtesy of Dr Krzysztof Ostaszewski
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Weekly Practice Exam "P" Problems courtesy of Dr Krzysztof Ostaszewski
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April 9 exercise
Here is another practice problem:
http://www.math.ilstu.edu/krzysio/495KOExercise.pdf
Yours,
Krzys'Last edited by admin; April 4 2005, 09:59 AM.Want to know how to pass actuarial exams? Go to: smartURL.it/pass
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Want to know how to pass actuarial exams? Go to: smartURL.it/pass
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April 23 exercise
Some discrete probability basics:
Enjoy.
Yours,
Krzys'Want to know how to pass actuarial exams? Go to: smartURL.it/pass
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question about the 3195 problem
I have a question about the 3195 problem posted. In the 4th line of the solution:
Cov(X1aX2bX3,X2) = Cov(X2,X1)  aVar(X2)  bCov(X3,X2)
I'm not sure how you got from left hand side to the right hand side of the equation. I've never seen an covariance example like this. I'm familiar with the equation for the covariance of 2 variables, but not of the variance of an expression and a variable. Could you generalize the rule for me? Thanks very much.
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Originally posted by engelbmjI have a question about the 3195 problem posted. In the 4th line of the solution:
Cov(X1aX2bX3,X2) = Cov(X2,X1)  aVar(X2)  bCov(X3,X2)
I'm not sure how you got from left hand side to the right hand side of the equation. I've never seen an covariance example like this. I'm familiar with the equation for the covariance of 2 variables, but not of the variance of an expression and a variable. Could you generalize the rule for me? Thanks very much.
Cov(x1  ax2  bx3, x2)
= Cov(x1,x2)  Cov(ax2,x2)  Cov(bx3,x2)
= Cov(x1, x2)  a*Cov(x2,x2)  b*Cov(x3, x2)
= Cov(x1, x2)  a*Var(x2)  b*Cov(x3, x2)
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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 17 2005, 02:21 AM.Want to know how to pass actuarial exams? Go to: smartURL.it/pass
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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'Want to know how to pass actuarial exams? Go to: smartURL.it/pass
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Exercise for April 30, 2005
I have now posted this exercise with a solution:
Enjoy.
Yours,
Krzys'Want to know how to pass actuarial exams? Go to: smartURL.it/pass
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Exercise for May 7, 2005
I have now posted this exercise with a solution:
Enjoy.
Yours,
Krzys'Want to know how to pass actuarial exams? Go to: smartURL.it/pass
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