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Old 09-15-2016, 01:06 PM
MathDoctorG MathDoctorG is offline
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Quote:
Originally Posted by RobertR1990 View Post
Q1: For #4 Why is g(x) monotone for f(x)= 2/(pi(1+4x^2))?
To my understanding of monotone, the function must be strictly increasing or decreasing. Not seeing this relationship because the first 3 terms plugging in -1,0,1 for x respectively I get [2/(pi(5)],[2/pi],[2/(pi(5)]). Which is not strictly increasing or decreasing. What am I misunderstanding?
You seem to be testing the original density of X for monotonicity. The issue isn't whether the density of the X is monotone, but rather whether or not the connecting function g such that Y = g(X) is monotone - that's the function that you require the inverse of, not f(x). This connecting function is just 1/x, which is monotone decreasing.


Quote:
Originally Posted by RobertR1990 View Post
Q2:For #21 Considering these random variables
X= Cost 1 follows Uniform(0,4)
Y= Cost 2 follows Uniform(0,4)
in thousands
P= Insurance Payment follows continuous distribution
P= { X + Y for 0<X+Y<6 , 6 for X+Y>6}
where P=min(X+Y,6000)
Based on the graph in the video I tried to split it into 2 separate regions above the line and below the line.
I: Below the line , II: Above the line
This is what I have: E(P)=([E(I)+E(II)]*(1/3)(1/3))
I think we might need two double integrals for the region below the line. Is that correct?
So for Region (I) I have: the outer integral from 2 to 4 and 0 to 6-x the inner integral (1/16)(X+Y) dy dx + the outer integral from 0 to 2 and 0 to 4 the inner integral (1/16)(X+Y) dy dx.

Region(II) = the outer integral from 6-x to 4 and 2 to 4 the inner integral (1/16)(6) dy dx.
Unfortunately I am not coming up with the correct answer perhaps this setup is incorrect? What am I misunderstanding?
So that's



Here's the integrations:
One part

Another part

Last bit

Now that's the expected payment in the case where both of the machines require maintenance.

If only one machine requires maintenance, then payment will cover the whole thing, since that amount won't exceed 6000. So that means we have an additional expected amount of 2000 with probability (1/3)(2/3) (case where the first machine breaks) and another 2000 with probability (2/3)(1/3) (case where just the second machine breaks.

The grand total, then is


And the difference is that I rounded the 0.435.


Quote:
Originally Posted by RobertR1990 View Post
Q3: For # 25 Why do we want the Var(min(X,Y)) and not Var(X+Y)?
Any help would be greatly appreciated. Thank you very much.
The question asks for the variance of the time until at least one of the servers has failed. At least one of the servers will have failed as soon as the first of them fails, which is the minimum of the two times until failure.

Does that help?
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