The key is note the possible values your payoff can take.
Do you agree with me that your payoff will only be valid for values of (ST> =0 to infinity) so your intergral will be from 0 to infinity. The formula on page 18 of the tables will give us a problem because log of 0 is invalid, although some clever people may make a certain assumption about the invalid Ln (0) and continue with the formula on pg18 of the tables but that assumption is quit confusing I won't talk about it here.
So after the invalid solution we see if we can use the formula on page 14 of the tables. Note that the values of x our lognormal distribution can take is (x>0) which match with (ST>0) and ST follows a lognormal distribution so we can use the rth moment of the lognormal distribution to evaluate E (ST^(1/2)).
Last edited by a moderator: Sep 13, 2019