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I am reading a paper and get a problem here, the following terms are all from standard BS models. the paper says using the well known fact $$Se^{-q(T-t)}N^{'}(d1)=Ke^{-r(T-t)}N^{'}(d2)$$ here the differentiation is respect to $T$.For instance $N^{'}(T^2+1)=2T$

So anyone could give some hints to get this fact?

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  • $\begingroup$ I suppose $q$ is a continuous yield dividend and $r$ the risk-free rate. But be careful, $N^\prime$ is most likely not the differential operator here! It is the probability density function of the standard normal. $\endgroup$
    – vanguard2k
    Apr 25, 2014 at 12:33

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The first equality is a bit tedious to derive but straight-forward. As commented by vanguard2k, the notation $N'(x)$ is meant to denote the density function of the standard normal distribution: $$ N'(x) = n(x) := \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} $$. Simply insert the $d_{i}$ terms, $$ d_1 = \frac{\log\left(\frac{S}{K}\right) + (r-q+\frac{1}{2}\sigma^2)(T-t)}{\sigma\sqrt{T-t}} $$ $$ d_2 = \frac{\log\left(\frac{S}{K}\right) + (r-q-\frac{1}{2}\sigma^2)(T-t)}{\sigma\sqrt{T-t}} $$ and rearrange.

The second equality cannot be, in general, correct. Simply write out the lhs to see that. Or note that the maximum of the standard normal pdf is slighlty less than 0.4. So for T>0.2, the equality is incorrect.

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  • $\begingroup$ Thx. Totally agreed about N'(x) $\endgroup$
    – Xun Bao
    Apr 26, 2014 at 0:29

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