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I read Higham's derivation of the Black-Scholes equation in "An Introduction to Financial Option Valuation". The issue I am having is that it relies on some assumptions related to a continuous time asset model, but these assumptions do not seem completely justified. I am hoping that someone has seen this derivation (but this may not be necessary to answer my question).

The idea is to start with a discrete time asset model that models the price $S(t)$ of an asset. The model is \begin{equation} S(t_{i+1}) = (1+\mu \delta t + \sigma \sqrt{\delta t} Y_i)S(t_i) \end{equation} where we are working on a time interval $[0,t]$ with $n$ subintervals $[t_i, t_{i+1}]$ of length $\delta t$. Here, $\mu$ and $\sigma$ are constants and $Y_i$ are I.I.D. random variables with distribution N(0,1). Now, when moving to the continuous time model (which corresponds to the limit $\delta t \to 0$), Higham writes \begin{equation} \text{log}\left(\frac{S(t)}{S(0)} \right) = \sum_{i=1}^n \text{log} \left(1 + \mu \delta t + \sigma \sqrt{\delta t} Y_i \right) \end{equation} and then expands the right hand side as a Taylor series to compute the expected value and variance of the left hand side. Appealing to the central limit theorem then allows one to write down a continuous time model for $S(t)$. The issue I have is with the $\text{log}$ on the right hand side. It seems that the argument can be negative and so this shouldn't be defined. How can one get around this issue? Higham also claims that the Taylor series expansion is valid even though the argument is a random variable. Is there also a way to make this statement rigorous?

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$\delta t \rightarrow 0$ as $n\rightarrow \infty$. In that limit jumping to 0 is probability 0 with a continuous diffusion model.

In the discrete case it is a numerical possibility but more of the artifact kind if the variance at each step is not too big.

On the other hand, if you simulate a jump diffusion model where for instance a bankrupcy probability is not 0 then you should take this possibility into account since $S_t$ is an absorbing state for the SDE.

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