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How is call delta mathematically derived from Black Scholes Model (without approximation) ? Please help me understand each step mathematically. And how it is approximated to say that delta is the probability of option expiring in the money?

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Look here for a detailed derivation of the formula for $\Delta$ (be aware that this particular website uses $r_d$ to denote the risk-free rate and $r_f$ to denote the dividend yield). You can always ask for more specific help regarding a particular step in the derivation.

It is easy to see that $\mathbb{Q}[\{S_T\geq K\}]= \Phi(d_2)$. Just replace $S_T=S_0\exp\left(\left( r-q-\frac{1}{2}\sigma^2\right)T +\sigma\sqrt{T}Z\right)$ where $Z\sim N(0,1)$ and isolate $Z$ on the left-hand side. This is the risk-neutral probability of expiring ITM. Note that $\Delta=\Phi(d_1)=\Phi(d_2+\sigma\sqrt{T})\approx \Phi(d_2)$. This is since $\sigma\sqrt{T}$ is typically very small.

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    $\begingroup$ That derivation is excellent but it starts with a small typo: in BSCallPrice the roles of $r_f$ the risk free rate and $r_d$ the dividend rate are reversed. $\endgroup$ – Alex C Oct 6 '19 at 15:39
  • $\begingroup$ I think it's not a typo because they use $r_f$ constantly and (as far as I checked) consistently to denote the dividend yield and $r_d$ as risk-free interest rate. So I believe they do this intentionally. However, I do agree with you that this notation is odd because $r_f$ should be the risk-free rate (or perhaps the rate in a foreign currcency) but not the dividend yield. $\endgroup$ – KeSchn Oct 6 '19 at 16:33
  • $\begingroup$ But I edited my answer and highlighted that notation issue. Thank you for the hint, Alex! $\endgroup$ – KeSchn Oct 6 '19 at 16:41
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    $\begingroup$ N represents the cumulative normal distribution, and n represents the normal density. Cumulative is integral of density, so density is then derivative of the cumulative - remember integral is anti-derivative, so derivative of anti--derivative is then the function. For technical approach, please see Leibniz rule (en.wikipedia.org/wiki/Leibniz_integral_rule). As @KeSchn outlined above, if we take derivative of N(s), we get n(s), and if we want to take derivative of N(d_1) where d_1 is a function of s, we will need to add chain rule: n(d1)*d(d1)/dS $\endgroup$ – Magic is in the chain Oct 6 '19 at 20:00
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    $\begingroup$ A reminder what N(.) and n(.) look like dwaincsql.files.wordpress.com/2015/05/normal-pdf-cdf-1.png $\endgroup$ – Alex C Oct 6 '19 at 22:42

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