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Past exam of the mathematics course of the University of Cambridge

2019

Paper 4

Given the present state $(S_t^1,M_t)=(x,m)$, take an independent fresh copy of the stock path and denote its initial value, terminal value after $T-t$ steps, and running maximum by $(y,z,w)=(S_0^1,S_{T-t}^1,M_{T-t})$. Scaling the fresh path to start at $x$ makes its terminal value $xz/y$ and its future running maximum $xw/y$. Including the maximum already attained gives
$$
M_T=\max\left\{m,\frac{xw}{y}\right\}.
$$
Hence define
$$
\boxed{g(x,m,y,z,w)=h\left(\frac{xz}{y},
\max\left\{m,\frac{xw}{y}\right\}\right).}
$$
Independent increments then give the <Running-maximum state reduction in the Cox--Ross--Rubinstein model>
$$
\boxed{V_t(\omega)=v_t(S_t^1(\omega),M_t(\omega)),}
$$
with
$$
\boxed{v_t(x,m)=\mathbb E_Q[g(x,m,S_0^1,S_{T-t}^1,M_{T-t})].}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 29K / d / Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 29K / d / Solution

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Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 29K / d / Solution

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