Trinomial model converges to Black-Scholes weakly

Question

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 1

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 $P(f(X)=f(x_i))=p_i$.