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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:

$\log(\frac{SN'(d_1)}{e^{-r(T-t)}EN'(d_2)}) = 0$

From the Black scholes formula:
$C(S,t)=SN(d_1)-Ee^{-r(T-t)}N(d_2)$

I've been puzzling arround but I'm stuck. This is where I came so far, do you know where I'm going wrong?

$\log(\frac{SN'(d_1)}{e^{-r(T-t)}EN'(d_2)}) = \log(SN'(d_1))-\log(e^{-r(T-t)}EN'(d_2))=0$

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  • $\begingroup$ $N'$ is the derivatve of the normal cdf - right? What if you plug in? $\endgroup$ – Richard Oct 22 '15 at 12:49
  • $\begingroup$ Yes, $N'$ is the derivative of a normal cdf. So you could rewrite that as $N'(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}$ But what do you mean with "plug in"? $\endgroup$ – Alfons Ingomar Oct 22 '15 at 13:55
  • $\begingroup$ Keep going with the substitution, you will see. $\endgroup$ – noob2 Oct 22 '15 at 14:03
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    $\begingroup$ You plug-in $d1$ and $d2$ ... $\endgroup$ – Richard Oct 22 '15 at 14:03
  • $\begingroup$ Please don't cross post your question, see this topic for guidance. $\endgroup$ – Bob Jansen Oct 26 '15 at 21:21
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I am trying to fill in what Richard left for the second part: \begin{align*} \exp(-r(T-t))E\, N'(d_2) &= \frac{1}{\sqrt{2\pi}}\exp(-r(T-t))E\, \exp\left(-\frac{1}{2}d_2^2\right) \\ &=\frac{1}{\sqrt{2\pi}}\exp(-r(T-t))E\, \exp\left(-\frac{1}{2}\big(d_1-\sigma\sqrt{T-t}\,\big)^2\right) \\ &=\frac{1}{\sqrt{2\pi}}\exp(-r(T-t))E\\ &\qquad\qquad \exp\left(-\frac{1}{2} d_1^2 -\frac{1}{2}\sigma^2 (T-t) + d_1 \sigma\sqrt{T-t}\right)\\ &=\frac{1}{\sqrt{2\pi}}\exp(-r(T-t))E\\ &\qquad\qquad \exp\left(-\frac{1}{2} d_1^2 +\ln(S/E) + r(T-t)\right)\\ &=\frac{1}{\sqrt{2\pi}} S \, \exp\left(-\frac{1}{2}d_1^2\right)\\ &= SN'(d_1). \end{align*} That is, \begin{align*} \ln\frac{SN'(d_1)}{\exp(-r(T-t))E\, N'(d_2)} = 0. \end{align*}

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  • $\begingroup$ +1...better with d1 and d2... $\endgroup$ – Richard Oct 23 '15 at 17:56
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The numerator is $$ S N'(d_1) = S \frac{1}{\sqrt{2 \pi}} \exp(-1/2 d_1^2) = \\ S \frac{1}{\sqrt{2 \pi}} \exp\left(- \frac12 \left(\log(S/E)+ (r + \frac12 \sigma^2(T-t)) \right)^2 / \sigma^2 (T-t) \right) $$ the denominator is: $$ \exp(-r (T-t)) E N'(d_2) = \\ E \frac{1}{\sqrt{2 \pi}} \exp\left(- \frac12 \left(\log(S/E)+(r- \frac12 \sigma^2(T-t)) \right)^2 / \sigma^2 (T-t) -r(T-t) \right). $$ Now what if we extend the square, match exp and log and if then nominator and denominator are equal we get the result.

EDIT: I did not finish the calculation. But the latex above is too much for a comment. Maybe you can do the calculation. It not that clear to see whether the claim is true...

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