Gamma of straddles

It might appear that differentiating with respect to $S$ once will give us a constant, and that differentiating with respect to $S$ twice will thus give us 0. However, since owning options always gives you long gamma, owning a straddle (which is a cal and a put) should give you long gamma.

What went wrong?

The issue here is that the straddle approximation only gives you the value of the straddle when the strike is equal to the ATM. Consider a 3-D graph where the x-axis is the underlying price and the y-axis is the strike of the straddle, and the z-axis is the price of the straddle. Then, all that we are given, is the z-values along the line $x = y$ , namely

$z ( x, x) = \frac{1}{2000} x \sigma \sqrt{t}$

The delta of the straddle is equal to $\frac{ \partial z(x, y)}{\partial x }$ , meaning that we should hold the strike fixed and let the underlying vary. Similarly, the gamma of the straddle is obtained as $\frac{ \partial^2 z(x) } { \partial x^2 }$ .

However, taking $\frac{ \partial z ( x, x) } { \partial x }$ would be equal to

$\frac{ \partial z ( x, y ) } { \partial x } |_{x = x} + \frac{ \partial z ( x, y ) } { \partial y } |_{y = x}.$

These effects cancel out (approximately), which is why we would obtain the value of 0.