sophieactrainee
Keen member
Hi there,
I am looking at Solution for Q7 of Sep 21 Paper A, where you are asked to prove that d < e^r < u, using proof by contradiction.
They break it up into two cases:
Case 1: e^r <= d < u
Case 2: d < u <= e^r.
However, in the notes they just look at case where e^r < d < u
Is it okay to just use the proof in notes, rather than going through all 2 cases in the ASET solutions?
I.e. would this proof be ok, instead of ASET solution?
Prove d < e^r < u:
Proof:
Assume otherwise, i.e. e^r < d < u
Then we could borrow £1*S0 of cash, and buy £1*S0 of stock
At time 0 this would be net cost of 0 , i.e. -1*S0 + 1*S0 = 0
Now at time 1 this would be worth
-e^4r S0+ S0*u = S0 ( u – e^r) > 0 since u > e^r
OR
-e^4r S0+ S0*d = S0 ( d – e^r) > 0 since d > e^r
So in either case they would get a positive payoff => abitrage opportunity => d < e^r < u holds.
Thanks
I am looking at Solution for Q7 of Sep 21 Paper A, where you are asked to prove that d < e^r < u, using proof by contradiction.
They break it up into two cases:
Case 1: e^r <= d < u
Case 2: d < u <= e^r.
However, in the notes they just look at case where e^r < d < u
Is it okay to just use the proof in notes, rather than going through all 2 cases in the ASET solutions?
I.e. would this proof be ok, instead of ASET solution?
Prove d < e^r < u:
Proof:
Assume otherwise, i.e. e^r < d < u
Then we could borrow £1*S0 of cash, and buy £1*S0 of stock
At time 0 this would be net cost of 0 , i.e. -1*S0 + 1*S0 = 0
Now at time 1 this would be worth
-e^4r S0+ S0*u = S0 ( u – e^r) > 0 since u > e^r
OR
-e^4r S0+ S0*d = S0 ( d – e^r) > 0 since d > e^r
So in either case they would get a positive payoff => abitrage opportunity => d < e^r < u holds.
Thanks