Devastators dumb questions

[Actuarialsuck]

I assume the integral in question is something like



Sure you could distribute and integrate the polynomial but it's easier to say then we know that .

Now we need to remember that the limits will change as well, before we went from x = a to x = b, but now u = x - c so now your bounds are from u = a - c to u = b - c so our integral would be

which is a lot easier to integrate.

[Devastator]

Thanks, this helps actuarialsuck!

I haven't seen calc since I last passed exam 100 in 1999. I'll post the example tomorrow (and in the future too). But it's very similar to what was posted above

[Devastator]

In the latest Sam, looking at the solution for quiz 51-1, when solving for a, where does the 1/16 come from? Thanks in advance!

[Actuarialsuck]

Originally Posted by Devastator

In the latest Sam, looking at the solution for quiz 51-1, when solving for a, where does the 1/16 come from? Thanks in advance!

Post it?

[Devastator]

Anyone help with soa 165? I do my deductibles a different way (don't really get their solution anyways)

So I'm using E(s)-E(s^3). E(s) = 2.8

For the second piece, I'm obv calculating expected value of the deductible. The ded pays 3 if there are 3 rolls or more. So I did:
3 * 1-p0-p1-p2.
Then for p0=0.
P1 = 2e^-2 *1.4
p2 = 2e-2 * 2.8.

2.8 - all that, I'm getting .693. Anyone tell me what I'm doing wrong please?

[oswaldcobblepot]

What section and lesson are you talking about?

[Devastator]

Originally Posted by oswaldcobblepot

What section and lesson are you talking about?

It's from the soa 289, problem #165

I'm sorry, that was dumb. I put asm. All this shit is getting to me.

[oswaldcobblepot]

Alright --- to do it your way, we agree that E[S] = 2.8.

The issue with your methodology is that you appear to be only considering losses based on the Poisson distribution, when in reality you need to consider aggregate losses, taking into account Poisson AND individual losses.

How would you have 0 aggregate losses? Poisson N=O: e^-2

How would you have 1 aggregate loss? Poisson N=1 AND individual loss 1: (2e^-2)(.6) = 1.2e^-2

How would you have 2 aggregate losses? Poisson N=1 AND one individual loss of 2 OR Poisson N=2 AND two individual losses of 1: (2e^-2)(.4) + (2^2)/2(e^-2)(.6^2) = 1.52e^-2

Now calculate E[S^3] = 0(e^-2) + 1(1.2e^-2) + 2(1.52e^-2) + 3(1-(e^-2)-(1.2e^-2)-(1.52e^-2)) = 2.0634

In closing, E[S] - E[S^3] = 2.8 - 2.0634 = .7365

[Devastator]

That is a great explanation, thank you very much!

[SliceApproximateIntegrate]

Originally Posted by sjb554

Once you do substitution 1,000 times, it starts to get easy. I am not sure about the problem mentioned above, but if its something like int[(x^4)*e^x], this is not that hard really...

u=x^4 dv=e^x
du=4x^3 v=e^x
du2=12x^2 v2=e^x
du3=24x v3=e^x
du4=24 v4=e^x
du5=0 v5=e^x

Then you just cross down switching between + and -....
u*v+du*v2+du2*v3+du3*v4+du4*v5 evaluated at the limits...

This is probably the hardest (most lengthy) substitution you will see. A small variation on this is when the du's continue infinitely, but there is an easy trick for solving that also.

I feel like the actually hard (confusing) substitutions come up more on differential equation type problems, which I have not yet encountered on this exam...


Can you post the integral you are talking about for those of us that are not using asm?

d00d that's actually integration by parts (not substitution). But if we're trying to make math more pink, we'd call it integration by s!

Although I never thought about it that way - calculating all the u's and dv's you'll need first and then just do the calculating part in one step. So thanks!