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Given that $S_0 = 1, u = \frac{5}{4}, d = \frac{4}{5}, r = \frac{1}{40}$:

The payoff of a digital option with a barrier B > S_0 on the running maximum is:

1 if $max\{S_0, ..., S_n\} \geq B$

0 if $max\{S_0, ..., S_n\} < B$

If we take n > 3 and $B = (\frac{5}{4})^3$ :

How do we show that the initial price of this option is:

$v_0 = \frac{1}{(1+r)^n}*P(M_n \geq 3)$

where M is the running maximum of a random walk Y i.e.

$Y_k = \sum_{i=1}^{k} ξ_i$ and $M_n = max\{0, Y_1,..., Y_n\}$?

I know that $ln(S_n) = ln(\frac{5}{4})\sum_{i=1}^{n} ξ_i$ where $ξ_i : \{+1, -1\}$ are i.i.d. with up probability = $\frac{1}{2}$ but I'm not sure how to incorporate this into the proof?

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