Why are the greeks for the underlying stock 0 with the exception of delta?

In my textbook that I am self-studying from it is given that (assuming the Black-Scholes framework):

1. $\Delta_{stock} = \partial S / \partial S = 1$
2. All other Greeks for the underlying stock = 0

I can see why $\Gamma_{stock} = 0$, from taking the partial derivative of $\Delta_{stock}$.

But why is there not some significance to $\theta_{stock}$ $\rho_{stock}$, $\psi_{stock}$, etc. and why are those necessarily 0?

It would seem to me that it could be useful taking the partial derivative of the stock price with respect to the risk free rate, the continuously compounded return on the stock, and the variance of the stock.

• What is your model ?? – user16651 Jun 28 '16 at 15:57
• Sorry, forgot to say - Black-Scholes – user2521987 Jun 28 '16 at 16:02
• Right, but my textbook is implying that would define $\Delta_{stock} = \frac{\partial S}{\partial S} = 1$, which would make $\Gamma_{stock} = \frac{\partial S^2}{\partial^2 S} = 0$...but why couldn't we do the same thing for the partial derivative of the stock with respect to $r, \delta, \sigma$, etc? – user2521987 Jun 28 '16 at 16:09
• What is your textbook and in which section? – Gordon Jun 28 '16 at 17:46
• Ask yourself this: you have an option position with a Theta of 3.14. How many shares of the stock do you need to buy (or sell) to bring the Theta to the value 2.718? You can't, because the option is going to expire regardless and holding some stock is not going to change that. That's why we say that a stock has no Theta (or zero Theta, if you prefer). The Theta comes from the fact tat "the clock is ticking" on the option. – Alex C Jun 29 '16 at 1:07