Actuarial Outpost normal distribution question SOA #64
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#1
07-03-2014, 11:50 PM
 lllj Member CAS Join Date: May 2011 Posts: 5,371
normal distribution question SOA #64

Getting confused about a negative sign here and it's bugging me.

σ=-.25

Normally if I'm looking for probability(x<#)
I do
(#-mean)/std = z, then N(z) to get probability

So in this case we want prob(lnx<u)=.95 so N(z)=.95 so z=1.645 so
1.645 = (u-1.075)/(-.25√2)
But then taking U=8*e^u that's giving me 13.1

So obviously I'm getting U and L switched somehow but I'm confused?

ugh how do i have a math degree? maybe just too late to be studying.

Last edited by lllj; 07-04-2014 at 12:08 AM..
#2
07-03-2014, 11:52 PM
 lllj Member CAS Join Date: May 2011 Posts: 5,371

or am i just being stupid and p(X>x) = N(z), not P(X<x)? i'm going to sleep
#3
07-03-2014, 11:53 PM
 jcmc2112 Member Join Date: Feb 2014 Posts: 1,448

How'd you get a negtive sigma? O_O
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#4
07-04-2014, 12:04 AM
 lllj Member CAS Join Date: May 2011 Posts: 5,371

says right in the solution

S(t) =8exp[0.5375t − 0.25Z(t)]

but i did
dY=2SdS+.5*2*(dS)^2
and dS=(α-δ)Sdt+σSdZ
so
dY=2S((α-δ)Sdt+σSdZ)+σ^2*S^2dt
dY/Y=2(α-δ+σ^2)dt+2σdZ

so comparing that to dY/Y=1.2dt-.5dZ,
2σ=-.5
σ=-.25

then i got m=(α-δ-.5σ^2)*2 by finding
2(α-δ+σ^2)=1.2=2(α-δ+(-.25)^2)
(α-δ)=.56875
m=(.56875-.5*(-.25)^2)*2=1.075
..right?

honestly i'm so sick of staring at this stuff though

Last edited by lllj; 07-04-2014 at 12:07 AM..
#5
07-07-2014, 10:35 PM
 phi349 Member SOA AAA Join Date: Jun 2013 Posts: 35

I had a question about this problem as well, but it was a different point. It's asking for the Higher confidence interval which we find as N^-1= (1-.90)/2 and equals (-1.64485, 1.64485). I then applied the lognormal confidence interval for stock prices by computing S(0)e^(a-d-.5(-.25)^2)*2+(-.25)*sqrt(2)*1.64485 (since it's asking for the higher lognormal interval). But in the solution provided by Actuarial Brew, they use the absolute value of the volatility to compute this. Does any one know why it was done this way? Thanks in advance!
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#6
07-07-2014, 10:36 PM
 phi349 Member SOA AAA Join Date: Jun 2013 Posts: 35

Quote:
 Originally Posted by jcmc2112 How'd you get a negtive sigma? O_O
It was provided in the problem as part of a GMB.
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#7
07-07-2014, 11:47 PM
 lllj Member CAS Join Date: May 2011 Posts: 5,371

Quote:
 Originally Posted by phi349 I had a question about this problem as well, but it was a different point. It's asking for the Higher confidence interval which we find as N^-1= (1-.90)/2 and equals (-1.64485, 1.64485). I then applied the lognormal confidence interval for stock prices by computing S(0)e^(a-d-.5(-.25)^2)*2+(-.25)*sqrt(2)*1.64485 (since it's asking for the higher lognormal interval). But in the solution provided by Actuarial Brew, they use the absolute value of the volatility to compute this. Does any one know why it was done this way? Thanks in advance!
maybe this is the same question as mine, since it seems to have to do with a negative sign
#8
07-08-2014, 09:45 PM
 lllj Member CAS Join Date: May 2011 Posts: 5,371

by the way just looked at this with a fresh set of eyes, still not making sense.

normal tables/calculators give P(z<N(z))
looking up 1.645 in a table gives P(z<1.645)=P(N(z)<.9)=P(X<x) (X variable, x constant)

z=(x-m)/s

for P(X<U)

1.645=(lnU-m)/s
1.645=(lnU-1.075)/(-.25√2)
U=e^(-1.645*.25√2+1.075)

why does this give me the lower bound helppp. clearly i'm getting something backwards, and i don't feel like this ever gives me a problem on anything besides this one problem.
#9
07-09-2014, 12:26 AM
 lllj Member CAS Join Date: May 2011 Posts: 5,371

mystery solved? i think the note on pg. 445 might answer my question.