Adigital option is one in which the payoff depends in a discontinuous way on the asset price. The simplest example is the cash-or-nothing option, in which the payoff to the holder at maturity T is X1{ST>K} where X is some prespecified cash sum. Suppose that an asset price evolves according to the binomial model in which, at each step, the asset price moves from its current value Sn to one of Snu and Snd. As usual, if T denotes the length of each time step, d < erT < u. Find the time zero price of the above option. You may leave your answer as a sum.

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Step-by-step explanation:

The Payoff to the holder at maturity T = X1{ST > K}, where X = pre-specified cash sum.

Current value of asset price = Sn  after evolvement of asset price in reference to binomial model

Hence at every  step value changes to SnU and Snd

where ; d < e^rΔT < u  and Δ T = length of each step

u = ups

d = downs

Determine the time zero price of the above option

First we will find the price after each period

After one period :

when the price goes up, Sn + ΔT = Su

when the price goes down, Sn + ΔT = Sd

After  period two

For Su

when price goes up, Sn +2ΔT = Su2

when price goes down, Sn +2ΔT = Sud

For Sd

when  price goes down, Sn +2ΔT = Sd2

when price goes up, Sn +2ΔT = Sdu

To get the time zero price we will have to apply the same procedure in reverse position den get the sum pf all ups and downs which is

sum of all ups and downs = ∑

∑  

1. 3.4/4

2. 0.8/7

3. 0.5 x 3

4. 0.9 x 6

5. 10.8

6. 2.35

7. 0.567

8. 9.35

9. 0.02

10. 0.06

11. 0.16

12. 0.42

13. 0.63

14. 5.6

15. 8.1

16. 9.9

17. 0.28

18. 0.48

19. 0.27

20. 0.49

21. 0.45

22. 0.72

23. 0.9

24. 0.88

25. 0.75

26. 0.16

27. 0.328

28. 0.495

29. 0.72

30. 1.068

let's cover over the formula of the vertex:

y = a(x – h)^2 + k

in this formula, (h,k) are the vertex, always remember that the positive or negative value of the vertex is based on the sign it has.

for example, if the vertex is -3,-6 then both the x and y value would have the positive sign attached to it and if it were a positive x and y value, then a negative sign would be attached to it.

with that being said, for our problem the x value is a positive one so we are looking for a number with a negative sign next to it, in this case the first and second choice.

however, for the first answer choice it isn't quite part of the formula so we know that the answer choice correctly displays the x value.

to confirm, we can see if the -1 fits correctly with the second answer choice, and as mentioned above a negative number will have a positive sign attached to it. so, for our problem we can see that -1 has been changed to +1, so we can confirm that the second answer choice is correct.

hence, our final answer is y=(x - 1)^2 + 1