Binomial Model Convergence Through Symmetric Probability and Subjective Return
Summary
This note examines conditions under which a binomial model’s scaled random increments converge to a standard normal distribution, connecting the model to the Black–Scholes framework. It presents two parameterizations: symmetric probabilities, where up and down moves are equally likely, and subjective return, where the move probabilities depend on the difference between the subjective drift and the risk-free rate. In both cases, the intended check is that the increment’s mean becomes negligible while its second moment approaches one.
The attempted symmetric-probability calculation verifies a zero mean and unit second moment for standardized up and down outcomes. An answer points out that a central limit theorem must account for the increment’s mean and variance, and that centering and scaling are necessary to obtain a standard normal limit. The note poses the corresponding derivation for subjective return but does not provide it. Its convergence conditions are stated without fully specifying the limiting assumptions or order notation, so the derivation needs care.
Key ideas
- A binomial price model can approximate Black–Scholes when its scaled increments satisfy suitable moment conditions.
- With symmetric probabilities, standardized up and down outcomes have zero mean and unit second moment.
- Subjective return changes the up and down probabilities to reflect the drift relative to the risk-free rate.
- A central limit argument must center increments by their mean and scale by their standard deviation.
- The note does not complete the subjective-return moment calculation.
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Full text
# Symmetric probability and subjective return
# Symmetric probability and subjective return
Let $\{Z_k\}_{k=1}^{N}$ be a sequence of i.i.d. random variables with the following distribution $$Z_k = \begin{cases} \alpha &\text{with probability} \ \hat{\pi}\\ -\beta &\text{with probability} \ 1 - \hat{\pi} \end{cases}$$ Then we have $$\ln(S_T) = ln(S_0) + (r - \frac{\sigma^2}{2})T + \sigma\sqrt{T}\frac{1}{\sqrt{N}}\sum_{k=1}^{N}Z_k$$ The only criteria for the convergence of binomial model to Black-Scholes model is that the random variables $Z_k$, $k = 1,\ldots,N$ must satisfy $\hat{\mathbb{E}}[Z_1] = o(\delta)$ and $\hat{\mathbb{E}}[Z_1^2] = 1 + o(1)$ i.e. $$\text{If} \ \hat{\mathbb{E}}[Z_1] = o(\delta), \ \text{and} \ \hat{\mathbb{E}}[Z_1^2] = 1+o(1), \ \text{then}$$ $$\frac{1}{\sqrt{N}}\sum_{k=1}^{N}Z_k \ \text{converges to} \ \mathcal{N}(0,1) \ \text{weakly}$$
Symmetric probability: $$u = \exp(\delta(r - \frac{\sigma^2}{2}) + \sqrt{\delta}\sigma), l = \exp(\delta(r - \frac{\sigma^2}{2}) - \sqrt{\delta}\sigma) , \ \text{and} \ \ R = r\delta$$ Then; $$\hat{\pi}_u = \hat{\pi}_l = \frac{1}{2}$$
Subjective return: $$u = \exp(\delta\nu + \sqrt{\delta}\sigma), l = \exp(\delta\nu - \sqrt{\delta}\sigma), \ \text{and} \ \ R = r\delta$$ Then; $$\hat{\pi}_u = \frac{1}{2}\left(1 + \sqrt{\delta}\frac{r - \nu - \frac{1}{2}\sigma^2}{\sigma}\right) \ \ \text{and} \ \ \hat{\pi}_l = \frac{1}{2}\left(1 - \sqrt{\delta}\frac{r - \nu - \frac{1}{2}\sigma^2}{\sigma}\right)$$
> Show $$\hat{\mathbb{E}}[Z_1] = o(\delta), \ \ \text{and} \ \ \hat{\mathbb{E}}[Z_1^2] = 1 + o(1)$$ in the following cases. a.) symmetric probability b.) subjective return
Attempted solution a.) $$E[Z_1] = 1\times \frac{1}{2} - 1\times\frac{1}{2} = 0$$ and $$E[Z_1^2] = 1^2\times\frac{1}{2} + (-1)^2\frac{1}{2} = 1$$
## Answer by M. Jeunesse (score 1, accepted)
https://quant.stackexchange.com/a/25258
your statement is quite imprecise.
See https://en.wikipedia.org/wiki/Central_limit_theorem
With :
- $(Z_k)_{k=1\dots n}$ i.i.d with $\mathbb{E}\left[Z_1\right] = \mu$ and $\text{Var}(Z_1)=\mathbb{E}\left[Z_1^2\right] -\mu^2=\sigma^2$
- and by denoting $\mathcal{N}(m,v)$ a normal variance with mean $m$ and variance $v$
we have : $$ \text{weak}\lim_{N\to\infty}\frac{1}{\sqrt{N}}\sum_{i=1}^n(Z_k-\mu) = \mathcal{N}(0,\sigma^2) $$ or alternatively $$ \text{weak}\lim_{N\to\infty}\frac{1}{\sigma\sqrt{N}}\sum_{i=1}^n(Z_k-\mu) = \mathcal{N}(0,1) $$ you can apply it straightforward to your problemShown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.