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I saw the following proof of theta in a paper I read, and I thought it looked pretty neat. Unfortunately I don't understand the step that they do. This is what they do:

enter image description here

Now, I don't get how they go from $S_0 n(d_1)\frac{\partial d_1}{\partial t} - Xe^{-rt}n(d_2) \frac{\partial d_2}{\partial t}$ to $S_0 n(d1) \frac{\partial (d_1-d_2)}{\partial t}$. Could anyone explain to me why this is true?

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    $\begingroup$ Consider the relationship of $d_1$ and $d_2$ as well as the relationship of $n(d_1)$ and $n(d_2)$. $\endgroup$
    – Gordon
    Oct 24, 2018 at 17:36
  • $\begingroup$ so $d_2 = d_1 - \sigma \sqrt{T}$, and consequently I thought $n(d_2) = n(d_1 - \sigma \sqrt{T})$. How can I use this to go further? $\endgroup$
    – John
    Oct 24, 2018 at 17:50
  • $\begingroup$ Would it help to write out the normal density function? $\endgroup$
    – John
    Oct 24, 2018 at 18:00
  • $\begingroup$ Yes, to write out the normal density function. $\endgroup$
    – Gordon
    Oct 24, 2018 at 18:02
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    $\begingroup$ See also this question. $\endgroup$
    – Gordon
    Oct 24, 2018 at 18:10

2 Answers 2

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There is a well known identity for the Black Scholes model: $S_0 n(d_1)-X e^{-rT} n(d_2) = 0$ (proof).

Using this allows you to combine these two terms:

$$S_0 n(d_1)\frac{\partial d_1}{\partial t} - Xe^{-rT}n(d_2) \frac{\partial d_2}{\partial t}$$

into

$$S_0 n(d1) (\frac{\partial d_1}{\partial t}-\frac{\partial d_2}{\partial t})$$

or

$$S_0 n(d1) \frac{\partial (d_1-d_2)}{\partial t}$$

Then we use the fact that $d_1-d_2=\sigma\sqrt{t}$

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Since Black Scholes Theta is for the Black–Scholes option pricing formula, the above step holds true.

For more info, refer page 3 and 4 of this pdf. http://moya.bus.miami.edu/~tsu/jef2008.pdf

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  • $\begingroup$ This paper just repeats the equations above with no further explanation. $\endgroup$
    – Alex C
    Nov 23, 2018 at 23:00

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