I just started working on the Black Scholes formula with help of the book Financial option valuation by Higham. Apparently you are possible to derive the following function:
log(SN′(d1)e−r(T−t)EN′(d2))=0
From the Black scholes formula:
C(S,t)=SN(d1)−Ee−r(T−t)N(d2)
I've been puzzling arround but I'm stuck. This is where I came so far, do you know where I'm going wrong?
log(SN′(d1)e−r(T−t)EN′(d2))=log(SN′(d1))−log(e−r(T−t)EN′(d2))=0
Answer
I am trying to fill in what Richard left for the second part: exp(−r(T−t))EN′(d2)=1√2πexp(−r(T−t))Eexp(−12d22)=1√2πexp(−r(T−t))Eexp(−12(d1−σ√T−t)2)=1√2πexp(−r(T−t))Eexp(−12d21−12σ2(T−t)+d1σ√T−t)=1√2πexp(−r(T−t))Eexp(−12d21+ln(S/E)+r(T−t))=1√2πSexp(−12d21)=SN′(d1).
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