Mapping Trinomial Increments to a Weak Black–Scholes Limit
Summary
The document explains how to identify the standardized random increment in a risk-neutral trinomial asset model and connect it to a weak convergence argument for Black–Scholes. Taking logarithms of each of the three possible one-step multipliers gives three corresponding values for the increment, with the original branch probabilities retained under this one-to-one transformation.
The proposed route is to check the increment’s mean and second moment, then apply the stated limit result for sums of independent increments. The questioner’s initial two-outcome setup is corrected because the model has three possible outcomes. The answer clarifies the mapping and probabilities, but does not carry out the moment calculations or prove convergence of the full price process; those steps are needed to complete the argument.
Key ideas
- The trinomial model yields three standardized increment values, one for each multiplier branch.
- A one-to-one transformation of outcomes preserves their probabilities.
- The weak-limit argument requires checking the increment’s mean and second moment as the time step shrinks.
- The response clarifies the setup but leaves the moment checks and process-level convergence proof to the reader.
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Full text
# Trinomial model converges to Black-Scholes weakly
# Trinomial model converges to Black-Scholes weakly
> Consider risk-neutral trinomial model with $N$ periods presented by $$S_{(k+1)\delta}H_{k+1}, \ \ \text{for} \ \ k=0,\ldots,N-1$$ where $\delta:=\frac{T}{N}$ and $\{H_k\}_{1}^{N}$ is a sequence of i.i.d random variables with distribution $$H_k = \begin{cases} e^{\delta(r-\sigma^2/2)+\sqrt{3\delta}\sigma} \ &\text{with probability} \ \hat{\pi} = \frac{1}{6}\\ e^{\delta(r-\sigma^2/2)} \ &\text{with probability} \ 1 - \hat{\pi} = \frac{2}{3}\\ e^{\delta(r-\sigma^2/2)-\sqrt{3\delta}\sigma} \ &\text{with probability} \ \hat{\pi} = \frac{1}{6}\\ \end{cases}$$ and $\hat{\pi}$ < 1/2. Show that as $\delta\rightarrow 0$, this trinomial model converges to the Black-Scholes model in the weak sense. Hint: Find $Z_k$ such that $\ln(H_k) = (r - \sigma^2/2)\delta + \sigma\sqrt{\delta}Z_k$. Then show (3.6)
(3.6) states that if $\hat{\mathbb{E}}[Z_1] = o(\delta)$ and $\hat{\mathbb{E}}[Z_1^2] = 1 + o(1)$, then $\frac{1}{\sqrt{N}}\sum_{k=1}^{N}Z_k$ converges weakly to $\mathcal{N}(0,1)$.
Attempted solution: Let $\{Z_k\}_{1}^{N}$ be a sequence of i.i.d. random variables with the following distribution $$Z_k = \begin{cases} \alpha \ &\text{with probability} \ \hat{\pi}\\ -\beta \ &\text{with probability} \ 1-\hat{\pi} \end{cases}$$ such that $\ln(H_k) = (r - \sigma^2/2)\delta + \sigma\sqrt{\delta}Z_k$.
I am not really sure what to do from here, any suggestions is greatly appreciated.
## Answer by M. Jeunesse (score 1, accepted)
https://quant.stackexchange.com/a/25299
I think you mix up marginal law, and law of the process.
Your $Z_k$ must have three values, you just have to write the value of $\ln(H_k)$ for each possible value
Let $X$ taking values $(x_1,x_2,...,x_n)$ and $p_i=P(X=X_i)$, then $P(f(X)=f(x_i))=\sum_{j=1}^n p_j\mathbf{1}_{f(x_j)=f(x_i)}$
if $f$ is a one-to-one mapping, you get $f(x_j)=f(x_i)\Rightarrow i=j$ and $P(f(X)=f(x_i))=p_i$.Shown 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.