3
$\begingroup$

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?

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

|improve this question
$\endgroup$
  • $\begingroup$ $N'$ 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
  • 1
    $\begingroup$ 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
2
$\begingroup$

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*}

|improve this answer
$\endgroup$
  • $\begingroup$ +1...better with d1 and d2... $\endgroup$ – Richard Oct 23 '15 at 17:56
1
$\begingroup$

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...

|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.