I am wondering what the term S means in the equation I have circled? I am not sure how to interpret it.
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From the GBM we have:
$$(S_{t}-S_{t-1}) \approxeq dS_t = \mu \cdot S_t \: dt + \sigma \cdot S_t \: dW_t,$$ 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 $dt\approx [t-(t-1)]=1$.
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$\begingroup$ Thank you so much! Makes complete sense now. $\endgroup$ – Shayan Slaesi Jan 12 at 5:36
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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.
