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Answer to Question #126964 in Finance for ELIZABETH RIMUI

Question #126964
ii) Determine the price of a European put option on a non-dividend paying stock when the stock price is sh. 69, the strike price is sh. 70, the risk-free rate is 5% per annum, the volatility is 35% per annum, and the time to maturity is three months?
1
2020-07-20T18:43:17-0400

"P(sh. 69, 0.25 years) = sh. 4.89"

Workings

A European put option gives the holder of the option the right, but not the obligation, to sell the underlying security at expiry date.

The Black- Scholes model is the best model to use in calculating the price of the put option in question. The model uses risk adjusted probabilities. The price of the European put option is found by the formulae:

"P(S, t) = Ke^{-rt}N(-d_{2})-SN(-d_{1})"

Where,

"d_{1} = \\dfrac {ln(\\dfrac{S}{K}) +(r + \\dfrac{\\delta^2}{2})t} {\\delta \\sqrt {t}}" ,

"d_{2} = d_{1} - \\delta \\sqrt {t}"

Where,

K is the option strike price,

N is the standard normal cumulative distribution function,

r is the risk-free interest rate,

S is the price of the underlying asset,

t is the time to option expiry, and,

"\\delta" is the volatility of the underlying asset.

It is given:

K = sh. 70,

S = sh. 69,

r = "\\dfrac {5\\%} {100\\%} = 0.05,"

"\\delta = \\dfrac {35\\%} {100\\%} = 0.35,"

t = "\\dfrac {3months} {12months} = 0.25 years"

Substitution and Simplification

"d_{1} = \\dfrac {ln(\\dfrac{69}{70}) +(0.05 + \\dfrac{0.35^2}{2})\u00d70.25} {0.35\\sqrt {0.25}}"

"d_{1} = \\dfrac {(-0.014388737) +(0.111250\u00d70.25)} {0.1750}"

"d_{1} = \\dfrac {0.013423763} {0.1750}"

"= 0.076707217"

Therefore,

"-d = -0.076707217" =−0.077 (to 3 decimal places)

"d_{2} = d_{1} - \\delta \\sqrt {t}"

"d_{2} = 0.076707217 - 0.35\u00d7\\sqrt {0.25}"

= 0.076707217 - 0.1750

= -0.098292783

Therefore,

"-d_{2} = 0.098292783"

"-d_{2} = 0.098 \\space (to \\space 3 \\space decimal \\space places)"

So,

"P(S, t) = Ke^{-rt}N(-d_{2})-SN(-d_{1})"

"P(sh. 69, 0.25years)"

"= 70e^{-(0.05\u00d70.25)}N(0.098)-69N(-0.077)"

"N(0.098) = \\phi(0.098)"

= 0.5391

"N(-0.077) = \\phi(-0.077)"

"= 1 - \\phi(0.077)"

= 1 - 0.5307

= 0.4693

"P(sh. 69, 0.25years) = 70e^{-0.0125}\u00d70.5391-69(0.4693)"

= sh. 37.268223457 - sh. 32.38170

= sh. 4.886523457

"= sh. 4.89"

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