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# 金融代写|金融微积分代写Financial Calculus代考|MATH7091 Backwards induction

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## 金融代写|金融微积分代写Financial Calculus代考|Backwards induction

In fact most of the hard work has already been done when we examined the branch model. Extending the results and intuitions of section $2.1$ to an entire binomial tree is surprisingly straightforward. The key idea is that of backwards induction – extending the construction portfolio back one tick at a time from the claim to the required starting place.

Consider, then, a general claim for our stock $S$. When we examined a single branching of our tree, we had the function $f$ dependent only on the node chosen at the end of a single tick period – here we can extend the idea of a claim to cover not only the value of $S$ at the time the claim is exercised but also the history of $S$ up until that point.
The tree structure of the stock was not entirely arbitrary – it embodies a one-to-one relationship between a node and the history of the stock’s path up to and including that node. No other history reaches that node; and trivially no other node is reached by that history. This is precisely that condition that allows us actually to associate a claim value with a particular end-node on our tree. We shall also insist on the finiteness of our tree. There must be some final tick-time at which the claim is fully determined. A condition not unreasonable in the real financial world. A general claim can be thought of as some function on the nodes at this claim time-horizon.

## 金融代写|金融微积分代写Financial Calculus代考|The two-step

We know that the expectation operator can be made to work for a single branch – here, then, we must wade through the algebra for two time-steps, three branches stuck together into a tree. If two time-steps work, then so will many.

Suppose that the interest rate over any branch is constant at rate $r$. Then there exists some set of suitable $q_j$ s such that the value of the derivative at node $j$ at tick-time $i, f(j)$, is
$$f(j)=e^{-r \delta t}\left(q_j f(2 j+1)+\left(1-q_j\right) f(2 j)\right)$$
That is the discounted expectation under $q_j$ of the time- $(i+1)$ claim values $f(2 j+1)$ and $f(2 j)$. So in our two-step tree (figure 2.4), the two forks from node 3 to nodes 6 and 7 , and from node 2 to nodes 4 and 5 , are both structurally identical to the simple one-step branch. This means that $f(3)$ comes from $f(6)$ and $f(7)$ via
$$f(3)=e^{-r \delta t}\left(q_3 f(7)+\left(1-q_3\right) f(6)\right) \text {, }$$
and similarly, $f(2)$ comes from $f(4)$ and $f(5)$, with
$$f(2)=e^{-r \delta t}\left(q_2 f(5)+\left(1-q_2\right) f(4)\right) \text {. }$$
Here $q_j$ is the probability $\left(s_j \exp (r \delta t)-s_{2 j}\right) /\left(s_{2 j+1}-s_{2 j}\right)$, so for instance
$$q_2=\frac{s_2 \exp (r \delta t)-s_4}{s_5-s_4}, \quad \text { and } \quad q_3=\frac{s_3 \exp (r \delta t)-s_6}{s_7-s_6} \text {. }$$

## 金融代写|金融微积分代写Financial Calculus代考|The two-step

$$f(j)=e^{-r \delta t}\left(q_j f(2 j+1)+\left(1-q_j\right) f(2 j)\right)$$

$$f(3)=e^{-r \delta t}\left(q_3 f(7)+\left(1-q_3\right) f(6)\right),$$

$$f(2)=e^{-r \delta t}\left(q_2 f(5)+\left(1-q_2\right) f(4)\right) .$$

$$q_2=\frac{s_2 \exp (r \delta t)-s_4}{s_5-s_4}, \quad \text { and } \quad q_3=\frac{s_3 \exp (r \delta t)-s_6}{s_7-s_6}$$

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