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D.W. Simpson and Company -- Actuary Salary Surveys |
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#1
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Can anyone explain question 8? thanks in advance, the solution to it is extremely confusing.
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#2
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They do explain that really weird. I think it's because SOA, doesn't like to use a normal of a negative number. This is my explanation:
S=S(0) and K=S(0)e^(rt). This means that PV(K)=S. Var[lnS(t)]=.4t, this implies that volatility=sqrt(.4) Because PV(K)=S, d1=(.4/2)*10/(2)=1. Because sqrt(.4*10)=2. d2=-1 Black-Scholes holds true, so Call = S(0)*N(d1)- S(0)*e^(rt)*e^(-rt)*N(d2) Call= S(0)* [N(d1)-N(d2)] Call= 100*[.8413-.1587] Call=68.26 I hope that helps. |
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#3
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Quote:
I had no idea that that equation implied the volatility. How did you solve for d1? i'm not sure I understand that either. do we just assume r = 0? |
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#4
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Quote:
ln[S(0)/(S(0)*e^(rt)]+rt =ln[S(0)/(S(0)*e^(rt)*e^(-rt)] = ln[S(0)*e^(rt)/(S(0)*e^(rt)] =ln[S(0)/(S(0)]+rt-rt. That's basically, four ways to write the beginning part of d1 and they all equal 0. We aren't assuming r=0, it's just that r gets cancelled, because PV(K)=S. Since that equals 0, d1=[0+(.4/2)*10]/[sqrt(.4)*sqrt(10)]=1. I hope that helps. |
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#5
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thank you very much bjz
for question 14, is it even necessary to mention that it is a straddle? i don't even know what it is... |
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#6
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It would be if we were expected to know the typs of options and what they are comprised of, and they didn't tell the payoffs |
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