# Black Scholes informal derivation - question about a term in the equation [closed]

I am wondering what the term S means in the equation I have circled? I am not sure how to interpret it.

$$(S_{t}-S_{t-1}) \approxeq dS_t = \mu \cdot S_t \: dt + \sigma \cdot S_t \: dW_t,$$ where $$W_t$$ denotes a Brownian motion. Here, the first part, $$(S_{t}-S_{t-1})$$, can be seen as a discretization of $$dS_t$$. Now, squaring the GBM, $$dS_t^2$$, is an informal way of denoting the quadratic variation of the process, that is $$[dS_t,dS_t]$$. A slightly informal derivation gives us: \begin{align} dS_t^2 & = [dS_t,dS_t]\\ &= [\mu \cdot S_t \: dt + \sigma \cdot S_t \: dW_t, \: \mu \cdot S_t \: dt + \sigma \cdot S_t \: dW_t]\\ &= \mu^2 S_t^2 [d_t, d_t] +2\cdot \sigma\mu S_t^2 [dt, dW_t] +\sigma^2 S_t^2 [dW_t,dW_t]\\ &= \sigma^2 S_t^2 \: dt, \end{align} where $$[dt,dW_t]=[dt,dt]=0$$ since any finite/bounded variation process (read deterministic function) has zero quadratic variation (I'm talking about the $$dt$$ term), and $$[dW_t,dW_t]=dt$$, since quadratic variation of two Brownian motions equals the time difference (you can also look at Ito's multiplication table for these results). From your above formulation, $$S = S_t$$, and denotes the spot price, and probably that $$dt\approx [t-(t-1)]=1$$.
It's the current spot price, or $$S_t$$. The "next chapter" might show if/why they drop the subscript for this approximation, but in the end the variable in the black-scholes equation will be the current spot price.