Proof Black Scholes Theta

Question

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?

Answer 1

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

Answer 2

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