# Proof Black Scholes Theta

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:

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?

• Consider the relationship of $d_1$ and $d_2$ as well as the relationship of $n(d_1)$ and $n(d_2)$. – Gordon Oct 24 '18 at 17:36
• 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? – Keith Oct 24 '18 at 17:50
• Would it help to write out the normal density function? – Keith Oct 24 '18 at 18:00
• Yes, to write out the normal density function. – Gordon Oct 24 '18 at 18:02
• See also this question. – Gordon Oct 24 '18 at 18:10

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 $$d_1-d_2=\sigma\sqrt{t}$$

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

• This paper just repeats the equations above with no further explanation. – Alex C Nov 23 '18 at 23:00