November 2012 Progress Thread

[lit041000]

go to bed son or your mother is going to get angry.

[go_for_it]

This is my first post on AO, although i have been visiting the site regularly for the past 7-8 months. I just sat for P (and passed) making it two exams under my belt. Next step will be MLC in November.

What do you think (or have been told) what is the approx. time needed to nail this "puppy"?

[MathDoctorG]

Originally Posted by lit041000

On page 48 of AMLCR, example 3.5, how do they justify expanding .7Q70.6 = .4Q70.6+(1-.4Q70.6).3Q71 ? They say as deaths are assumed to be uniform but I don't follow

That statement should be true regardless of the model. The equation just states that the probability of dying in the next .7 of a year, having survived to age 70.6, is the same as the probability of dying in the next .4 of a year, having survived to age 70.6, plus the probability of surviving during that .4 of a year and then dying in the next .3 of a year after that. But surviving that .4 of a year is 1 minus the probability of not surviving during that time, or
,
and then the probability of dying in the .3 years after that, since the individual has now survived to age 70.6+.4, is
.
You multiply these because this is an intersection of events - survival to age 71 AND death in the next .3 after age 71, and these events are typically assumed to be independent.

[NBran]

Originally Posted by lit041000

On page 48 of AMLCR, example 3.5, how do they justify expanding .7Q70.6 = .4Q70.6+(1-.4Q70.6).3Q71 ? They say as deaths are assumed to be uniform but I don't follow

You really need to learn TeX on the MLC forum

[lit041000]

Originally Posted by MathDoctorG

That statement should be true regardless of the model. The equation just states that the probability of dying in the next .7 of a year, having survived to age 70.6, is the same as the probability of dying in the next .4 of a year, having survived to age 70.6, plus the probability of surviving during that .4 of a year and then dying in the next .3 of a year after that. But surviving that .4 of a year is 1 minus the probability of not surviving during that time, or
,
and then the probability of dying in the .3 years after that, since the individual has now survived to age 70.6+.4, is
.
You multiply these because this is an intersection of events - survival to age 71 AND death in the next .3 after age 71, and these events are typically assumed to be independent.

I am going to have investigate why surival and death of two mutually exclusive time intervals is independent when the probability is formulated as a conditional A | B. And then how the event of dying in the next .7 time can be decomposed into the sum. I know that is the case of the sum of two mutually exclusive events that cover the entire probability space but I don't yet see how those two events cover it. I will keep digging with this help. My best guess is that I've forgotten most of the identities from P and need to dig

[Gandalf]

Originally Posted by lit041000

I am going to have investigate why surival and death of two mutually exclusive time intervals is independent when the probability is formulated as a conditional A | B. And then how the event of dying in the next .7 time can be decomposed into the sum. I know that is the case of the sum of two mutually exclusive events that cover the entire probability space but I don't yet see how those two events cover it. I will keep digging with this help. My best guess is that I've forgotten most of the identities from P and need to dig

"Independent" is not really the right term, since death in interval one and death in interval 2 are mutually exclusive events.

This formula, though, is true from Course P: P(A and B)=P(A)*P(B|A).

Can be applied with A="Surviving interval 1".

[MathDoctorG]

Originally Posted by Gandalf

"Independent" is not really the right term, since death in interval one and death in interval 2 are mutually exclusive events.

This formula, though, is true from Course P: P(A and B)=P(A)*P(B|A).

Can be applied with A="Surviving interval 1".

Right, thanks for the clarification. Also, there is a discussion in AMLCR on page 18 and 19 regarding this issue.

[lit041000]

Originally Posted by MathDoctorG

Right, thanks for the clarification. Also, there is a discussion in AMLCR on page 18 and 19 regarding this issue.

I couldn't ask for better help from the forum.

I follow now. We partition the probability space ( dying in .7 years ) into two mutually exclusive events ( dying within .4 years) and (surviving the first .4 years).

Then you get to use the following equations



it's been a while since I've had to think through soft logic like this. the last test I studied for didn't have as much wiggle room or 'open space' in the inferences if you follow my meaning.

[Bama Gambler]

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[GIFADE-208]

I started studying MLC today. There's gonna be little fun this summer; but it's worth it. I'm using TIA and ACTEX...plus I got some text books for referencing. I wish everybody the best in here and I hope we have discussions that increase our odds of success in Fall.