I'm constructing a binomial (decision) tree with nodes $ x_i $ according to the following recursion:

$ x_{i+1} = \begin{cases} x_i e^{\mu \delta t + \sigma(x_i)\sqrt{\delta t}} & \text{for next up node} \\ x_i e^{\mu \delta t - \sigma(x_i)\sqrt{\delta t}} & \text{for next down node} \end{cases} $


$ \sigma (x_i) = \sigma_0\left( 1 - \alpha \frac{x_i - x_0}{x_0} \right) $

Taking the limit of $x_{i+1}$ for $\delta t \rightarrow 0$, I get:

$ \lim\limits_{\delta t \rightarrow 0} x_{i+1} = \begin{cases} \lim\limits_{\delta t \rightarrow 0} x_i e^{\mu \delta t + \sigma(x_i)\sqrt{\delta t}} = x_i & \text{for next up node} \\ \lim\limits_{\delta t \rightarrow 0} x_i e^{\mu \delta t - \sigma(x_i)\sqrt{\delta t}} = x_i & \text{for next down node} \end{cases} $

Therefore, $\lim\limits_{\delta t \rightarrow 0} x_{i+1} = x_i $. Using this limit recursively, I yield: $\lim\limits_{\delta t \rightarrow 0} x_{i+1} = x_0 $. Therefore:

$ \lim\limits_{\delta t \rightarrow 0} \sigma (x_i) = \lim\limits_{\delta t \rightarrow 0} \sigma_0\left( 1 - \alpha \frac{x_i - x_0}{x_0} \right) = \sigma_0\left( 1 - \alpha \frac{x_0 - x_0}{x_0} \right) = \sigma_0 $

Is my notation correct? Is this last expression correct?

Am I allowed to take the limit recursively to yield $\lim\limits_{\delta t \rightarrow 0} x_{i+1} = x_0 $ ?


What you are trying to express is taking a limit of functions, and the notation has confused you. If I am not mistaken, this is a binomial tree.

Let $n$ be an integer. I believe what you mean to express is the sequence of functions $\{x^{(n)}\}$ with

\begin{gather*} x^{(n)}((k+1)\delta t)=x^{(n)}(k\delta t)\exp(\mu\delta t+Y_{k}^{(n)}\sigma(x^{(n)}(k\delta t))\sqrt{\delta t})\\ \text{where }\delta t=T/n\text{ and }0\leq k<n\text{ is an integer}. \end{gather*} $Y_{k}^{(n)}$ denotes a random variable with support $\{-1,+1\}$. $x^{(n)}(0)$ is a constant independent of $n$.

When $t/\delta t$ is not an integer, $x^{(n)}(t)$ is defined by linear interpolation. The limit you care about is $$ \lim_{n\rightarrow\infty}x^{(n)}(t). $$ Note that $n$ depends on $\delta t$. Another subtlety is that this is not a limit in the usual sense, since randomness is involved (look up "convergence in distribution").

P.s. ~ This teaching note might be of interest.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.