1. Let S = {S(t) | t ≥ 0} be the price process of some...

1. Let $S = \{S(t) | t \ge 0\}$ be the price process of some stock. Let $C(t)$ denote the value at time $t$ of a European call option written on $S$ with maturity date $T$ and exercise price $K$. Then $C(T) = \max[0, S(T)-K]$. Draw a graph, plotting $C(t)$ versus $S(t)$. This is called the payoff graph for the this call option. The profit graph is the plot of $C(T)-C(0)$ versus $S(T)$. Draw the profit graph. For what values of $S(T)$ will the profit be positive ? (4 Marks)

Transcript

00:01 Let nx be cumulative function for normal distribution and this value is given by nx is equal to integral minus infinity to x e to the power minus t square by 2 upon root 2 pi dt.

00:41 On putting limit extending to infinity nx we will get minus infinite to infinite e to the power minus t square by 2 upon root 2 pi dt and this will be equal to root 2 pi upon root 2 pi and this value will be equal to 1.

01:07 Now let us consider the first case here.

01:12 Case 1 when st is greater than k.

01:18 So on taking log here ln st by k will be greater than 0.

01:29 Now taking the function that is given in the question limit t tending to capital t log st by k plus r plus 1 upon 2 sigma square capital t minus t upon sigma square root of t minus t and that is equal to 0.

01:58 Now when t is tending to capital t then d1 will be tending to infinite.

02:05 Similarly when t is tending to capital t d2 will also tend to infinite.

02:13 Now since the function the cumulative function nx is continuous therefore when t is tending to infinity then when t is tending to capital t then nx will tend to infinite.

02:38 Now the value of ct will be st into 1 minus e raised to the power 0 k into 1 and this as as t will tend to capital t.

02:54 So from here we will get the value of ct as st minus k and given the condition that t is tending to capital.

03:06 In the second case we will take st equal to k.

03:15 Now log value st upon k will be equal to 0.

03:21 So the value of d1 will be given by d1 is equals to r plus 1 by 2 sigma square capital t minus t upon sigma root t minus t...

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