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What is the smallest information structure that is required for using the binomial tree to calculate the price of a barrier (up-and-in) option? My gut feeling is any node below the node that reaches the barrier price will be irrelevant.

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    $\begingroup$ You would price the option as an up-and-out option with in-out-parity. In that case, the price at any node above the barrier is just 0 and you don't have to construct the corresponding part of the tree beyond the first such node. By the way - you could have edited your old question quant.stackexchange.com/questions/43820 instead of deleting it. $\endgroup$
    – LocalVolatility
    Jan 31, 2019 at 21:58
  • $\begingroup$ Thanks, guys. I appreciate your help. I will continue from there. $\endgroup$
    – veryBigman
    Jan 31, 2019 at 22:04

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I'm assuming you're talking about a European option. I did a similar problem for my homework recently, I used the in-out parity for pricing the up and in barrier option.

Basically European Option = Knock up and in Option + Knock up and out option

You can price the up and out easily using Binomial and use BS formula for pricing the European Option, then use the above parity to get the knock up and in.

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  • $\begingroup$ Why does calculating up and out require less information than calculating up and in? I mean, if in most of the state, the option is out, is it more convenient to use the binomial tree to calculate up and in straightway? $\endgroup$
    – veryBigman
    Jan 31, 2019 at 22:15
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    $\begingroup$ Because for an up and out if it is above the barrier, we can surely assign the price zero. But for an up and in, if it's below the barrier we can't assign the price zero cause it may have crossed the barrier some time ago $\endgroup$
    – user23564
    Jan 31, 2019 at 22:18
  • $\begingroup$ This is assuming we start from the Expiry time and start back inducting the prices upto time =0.. I'm not sure if I'm able to put it across clearly but I ran into this issue and hence did it this way $\endgroup$
    – user23564
    Jan 31, 2019 at 22:20
  • $\begingroup$ very clear , i understood. thx $\endgroup$
    – veryBigman
    Jan 31, 2019 at 22:30

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