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 Probability Old Exam P Forum

#31
03-09-2013, 01:38 PM
 007NewHere Member Non-Actuary Join Date: Feb 2013 Location: Mars Studying for Unknown College: Senior Posts: 805

From #25 of SE4.........I have a question about Uniform...

If given X & Y are uniformly distributed on a region, what about X or Y separately...can we assume that they are both uniform or not???

For #25, I computed it by using f(x,y)=f(x)*f(y)=1/16.......the result is 0.56.
So I don't know if I did computation wrong or I used wrong f(x,y) at the beginning???
#32
03-09-2013, 01:47 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 31,286

Quote:
 Originally Posted by 007NewHere From #25 of SE4.........I have a question about Uniform... If given X & Y are uniformly distributed on a region, what about X or Y separately...can we assume that they are both uniform or not??? For #25, I computed it by using f(x,y)=f(x)*f(y)=1/16.......the result is 0.56. So I don't know if I did computation wrong or I used wrong f(x,y) at the beginning???
If the region is a square or rectangle (and the edges are parallel to the x and y axes), yes, x and y are each uniform. Otherwise, x and y are not uniform. (In special cases one of them might be uniform, but often neither would be uniform, and I don't think it would be possible for both to be uniform.)
#33
03-09-2013, 02:54 PM
 007NewHere Member Non-Actuary Join Date: Feb 2013 Location: Mars Studying for Unknown College: Senior Posts: 805

Quote:
 Originally Posted by Gandalf If the region is a square or rectangle (and the edges are parallel to the x and y axes), yes, x and y are each uniform. Otherwise, x and y are not uniform. (In special cases one of them might be uniform, but often neither would be uniform, and I don't think it would be possible for both to be uniform.)
They look like uniform...but not real...that's why I did wrong...

Thank you!!!
#34
02-12-2014, 12:35 PM
 sluggerbroth Non-Actuary Join Date: Sep 2013 Posts: 9
TIA?

I will be prepping for P exam later this year. Not sure which is best way to prep. Does TIA have written material that can be studied at my pace? Are the video's offered only for a period of time? How does TIA compare ASM? BPP?
#35
12-10-2014, 02:27 PM
 Whenindoubtcalculate SOA Join Date: Dec 2014 Location: Utah, Orem Studying for P College: Utah Valley University Posts: 14

ln dx dx for variance of mgf is my favorite. a favorite for TIA as well unless they want us to find the second moment.
#36
09-15-2016, 11:27 AM
 RobertR1990 Member SOA Join Date: Jul 2015 Studying for Studying for P College: Binghamton University Alumni Posts: 54
TIA Practice Exam 2 Questions

Hello. I am wondering if someone could help me out with a few problems such as #4 and #21, #25 respectively on TIA practice exam 2.

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?

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?

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.
#37
09-15-2016, 12:06 PM
 MathDoctorG Member Non-Actuary Join Date: Nov 2010 Location: The 'ville College: Cornell Posts: 654

Quote:
 Originally Posted by RobertR1990 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 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 06} 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

$E(P\mid \text{both break)\cdot P[\text{both break}] = ((3/2+5/3) + 0.75)(1/3)(1/3) = 47/108 \approx 0.435$

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

$0.435 + 2\cdot (1/3)(2/3) + 2 \cdot (2/3)(1/3) \approx 1.324$
And the difference is that I rounded the 0.435.

Quote:
 Originally Posted by RobertR1990 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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#38
09-15-2016, 12:24 PM
 RobertR1990 Member SOA Join Date: Jul 2015 Studying for Studying for P College: Binghamton University Alumni Posts: 54
TIA Practice Exam 2

Yes it does MathDoctorG thank you very much.
#39
09-17-2016, 09:19 PM
 RobertR1990 Member SOA Join Date: Jul 2015 Studying for Studying for P College: Binghamton University Alumni Posts: 54
TIA Practice Exam 3 #24

Hello I think I am misunderstanding how to organize the payment structure for this problem. I hope someone could help me out.
N= # of Losses ~Poisson(lamda=2.5)
Y= Payment
Goal : E(Y)
I understand there are 2 blocks of payments; first block is a constant 5000; the second block varies; so in this case for the first loss we have (5000(for block 1) and 5000(for block 2) representing the 10000 in total for the first loss, the second loss we have (5000 for (block 1) , 2500 for (block 2))representing the 7500 in total for the second loss etc.
Q: So how do we use those terms representing the 2 blocks of payments regarding to the expectation?
Any help would be greatly appreciated. Thank you very much.

Last edited by RobertR1990; 09-18-2016 at 11:22 AM..
#40
09-19-2016, 12:41 PM
 MathDoctorG Member Non-Actuary Join Date: Nov 2010 Location: The 'ville College: Cornell Posts: 654

We include the 5000 second block whenever N is greater than 1, since in that case we know that the additional 5000 was paid. We include the 2500 second block when N is at least 2, since we know that in that case the additional 2500 was paid.

Try it directly.

$E[P] = 10000P[N=1] + 17500P[N=2] + 22500P[N=3] + \ldots$

Rewrite as an increasing bit and some constant bits:

$E[P] = 5000 P[N=1] + 5000P[N=1] + 10000P[N=2] + 5000P[N=2] + 2500P[N=2] + 15000P[N=3] + 5000P[N=3] + 2500P[N=3] + \ldots$

$= (\sum 5000nP[N=n]) + 5000\sum_{n=1} P[N=n] + 2500 \sum_{n=2} P[N=n] = E[N] + 5000P[N\ge 1] + 2500 P[N\ge 2]$

Does that help?
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