**CppQuant Answer**

- d1 = ( ln(S0/X) + (r+stderr^2/2)*T) / (stderr*T^1/2) = ( ln(100/100) + (0.06 + 0.12/2)*1) / (0.1*1^(1/2)) = 0.65
- d2 = d1 - (stderr*T^1/2) = 0.65 - (0.1*1^(1/2)) = 0.55
- N(d1): In Excel, Normsdist(0.65) = 0.74
- N(d2) = Normsdist(0.55) = 0.71
- C = S(0) *N(d1) - X * e^(-r*t) * N(d2) = $100 x 0.74 - $100 x e^(-0.06 x 1) x 0.71 = $7.46
- P = X*e^(-r*T)*(1-N(d2)) - S(0)*(1-N(d1)) = $100*e^(-0.06*1) * (1-0.7088) - $100*(1-0.7422) = $1.64.

*Bonus Points*

- Normsdist finds the cumulative normal values. ie. Normsdist(0.65) is the probability that a normally distributed variable with a zero mean and a standard deviation of 1.0 will have a value equal to or less than the 0.65.
- Model Assumptions
- The underlying price follows a lognormal probability distribution as it evolves through time.
- Reasonable for most assets that offer options. Additionally, the variance of the return is assumed to be constant for the life of the option.
- Interest rates remain constant and known.
- Problematic for pricing options on bonds and interest rates.

- The volatility of the underlying asset is known and constant.
- Specified in the form of the standard deviation of the log return.

- No transaction costs or taxes.
- No cash flows on the underlying.
- European exercise terms are used: not a major concern because very few calls are ever exercised before the last few days of their life.
- True because when you exercise a call early, you forfeit the remaining time value on the call and collect the intrinsic value. Towards the end of the life of a call, the remaining time value is very small, but the intrinsic value is the same.

- The underlying price follows a lognormal probability distribution as it evolves through time.

*Category: C++ Quant > Derivatives > Valuation*

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