## Pricing Down-and-in call options

This is a knock in barrier option. The option comes into existent only after the underlying’s price crosses a certain barrier price, H. The barrier lies below the underlying’s price at inception, hence the “down” in the title above.

Unlike the previous options discussed we cannot simply work backwards down the tree from terminal nodes to inception. This is because the value at any particular node depends on how the node was reached, that is whether the underlying price had crossed the barrier prior to the node or not. If the barrier was crossed prior to reaching the node the option is in effect if not it is still not “alive”.

Hence at every node there are two possible outcomes and therefore this requires the construction of two separate spread sheets/ tables. The first table is for option values that assume that the option has been knocked in prior to the current node. The procedure, therefore, is the same as the basic European call procedure. The value of the option if previously knocked in obtained from this table will be labelled **B _{+}**.

_{ }_{ }

The second table assumes that the option has not been knocked in previously. It would require the following adjustments:

**Step2**: The terminal payoffs will be

- if S>H, 0
- if S?H, max{0,S
_{T}-K}

**Step3**: The intrinsic values at non-terminal nodes is given below for times T-1?t and T-2?t: (i.e. for n=2)

Column | A | B | C | D |

Row | S_{T} |
C_{T} |
T-1?t | T-2?t |

1 |
=u^{2}S_{0} |
=if[A1>H,0,max(A1-K,0)] | =if[and(B1=0,A1<=H), corresponding cell in 1^{st} table, exp(-r?t)*{p*B1+(1-p)*B2}] |
=if[and(C1=0,A1<=H),corresponding cell in 1^{st} table, exp(-r?t)*{p*C1+(1-p)*C2}] |

2 |
=A1*d | = if[A2>H,0,max(A2-K,0)] | = if[and(B2=0,A2<=H), corresponding cell in 1^{st} table, exp(-r?t)*{p*B1+(1-p)*B3}] |
= if[and(C2=0,A2<=H), corresponding cell in 1^{st} table, exp(-r?t)*{p*C1+(1-p)*C3}] |

3 |
=A2*d | = if[A3>H,0,max(A3-K,0)] | = if[and(B3=0,A3<=H), corresponding cell in 1^{st} table, exp(-r?t)*{p*B2+(1-p)*B4}] |
= if[and(=H), corresponding cell in 1^{st} table, exp(-r?t)*{p*C2+(1-p)*C4}] |

4 |
=A3*d | = if[A4>H,0,max(A4-K,0)] | = if[and(B4=0,A4<=H), corresponding cell in 1^{st} table, exp(-r?t)*{p*B3+(1-p)*B5}] |
= if[and(C4=0,A4<=H), corresponding cell in 1^{st} table, exp(-r?t)*{p*C3+(1-p)*C5}] |

5 |
=A4*d | = if[A5>H,0,max(A5-K,0)] | = if[and(B5=0,A5<=H), corresponding cell in 1^{st} table, exp(-r?t)*{p*B4+(1-p)*B5}] |
= if[and(C5=0,A5<=H), corresponding cell in 1^{st} table, exp(-r?t)*{p*C4+(1-p)*C5}] |

Figure 95: Formulas for spreadsheet implementation of binomial tree for down and in call options assuming option not previously knocked in

The adjusted formulas in column C will then be copied to D and later columns depending on the number of steps employed. The value of the option if not previously knocked in obtained from this table will be labelled **B _{–}**.

The price of the option will then depend on the underlying price at inception. If S_{0}>H then the price of the option is **B _{–}**. Alternatively if S

_{0}?H, the price would be

**B**

_{+.}For the following parameters the numerical example is given below:

n | 4 | T | 0.1 | x | 0.998751 |

r | 0.05 | ?t | 0.025 | K | 50 |

? | 0.3 | d | 0.953673 | q | 0.498677 |

u | 1.048577 | p | 0.501323 | S_{0} |
47 |

The Barrier is set at H=45.

The first spread sheet representing that the option has already been knocked in is as follows:

Column | A | B | C | D | E | F |

Row | S_{T} |
C_{T} |
T-1?t | T-2?t | T-3?t | T-4?t |

1 |
56.820 | 6.820 | 5.500 | 4.871 | 4.330 | 4.014 |

2 |
54.187 | 4.187 | 4.250 | 3.798 | 3.707 | 3.376 |

3 |
51.677 | 1.677 | 2.097 | 2.546 | 2.425 | 2.595 |

4 |
49.283 | – | 0.840 | 1.050 | 1.484 | 1.476 |

5 |
47.000 | – | – | 0.420 | 0.526 | 0.848 |

6 |
44.823 | – | – | – | 0.211 | 0.263 |

7 |
42.746 | – | – | – | – | 0.105 |

8 |
40.766 | – | – | – | – | – |

9 |
38.877 | – | – | – | – | – |

Figure 96: Spreadsheet for down and in call options assuming option previously knocked in

The value of the option if previously knocked in, **B _{+ }**is equal to 0.848.

The second spread sheet representing that the option has not previously been knocked in is as follows:

Column | A | B | C | D | E | F |

Row | S_{T} |
C_{T} |
T-1?t | T-2?t | T-3?t | T-4?t |

1 |
56.820 | – | – | – | – | – |

2 |
54.187 | – | – | – | – | – |

3 |
51.677 | – | – | – | – | – |

4 |
49.283 | – | – | – | – | – |

5 |
47.000 | – | – | – | – | 0.105 |

6 |
44.823 | – | – | – | 0.211 | – |

7 |
42.746 | – | – | – | – | 0.105 |

8 |
40.766 | – | – | – | – | – |

9 |
38.877 | – | – | – | – | – |

Figure 97: Spreadsheet for down and in call options assuming option not previously knocked in

The value of the option if not previously knocked in, **B _{– }**is equal to 0.105.

The price of the option is 0.105 as S_{0 }=47 > H=45.

This overall coetnnt is fantastic. You make many points that I can agree with and others I need to ponder a bit. Your information coetnnt and writing ability is well above my expectations. Great work!