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#1




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 = (u1.075)/(.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; 07032014 at 11:08 PM.. 
#3




How'd you get a negtive sigma? O_O
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#4




says right in the solution
S(t) =8exp[0.5375t − 0.25Z(t)] but i did dY=2SdS+.5*2*(dS)^2 and dS=(αδ)Sdt+σSdZ so dY=2S((αδ)Sdt+σSdZ)+σ^2*S^2dt dY/Y=2(αδ+σ^2)dt+2σdZ so comparing that to dY/Y=1.2dt.5dZ, 2σ=.5 σ=.25 then i got m=(αδ.5σ^2)*2 by finding 2(αδ+σ^2)=1.2=2(αδ+(.25)^2) (αδ)=.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; 07032014 at 11:07 PM.. 
#5




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^(ad.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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#7




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#8




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=(xm)/s for P(X<U) 1.645=(lnUm)/s 1.645=(lnU1.075)/(.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. 
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