First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 2. SOLUTIONS OF EQUATIONS: ROOT-FINDING 68
Finally, we have the derivative we need:
dC
d 2
dσ = Se− 1
2
√
2π
(
√T log( S σ2
)+(r + )T
−
K 2
σ 2√ T
) ( d 2
+ K e−(rT+ 2
2 ) log(
S
σ2
)+(r +
K
√ )T
)
2
2π
σ 2√ T
(2.10)
We are ready to apply Newton’s method to solve the equation C(σ) − Ĉ =0. Now let’s
find some data.
The General Electric Company (GE) stock is $7.01 on Dec 8, 2018, and a European call
option on this stock, expiring on Dec 14, 2018, is priced at $0.10. The option has strike price
K =$7.5. The risk-free interest rate is 2.25%. The expiry T is measured in years, and since
there are 252 trading days in a year, T =6/252. We put this information in Julia:
In [1]: S=7.01
K=7.5
r=0.0225
T=6/252;
We have not discussed how to compute the distribution function of the standard
∫
normal random variable φ(x) = √ 1 x /2
2π −∞ e−t2 dt. In Chapter 4, we will discuss how to
compute integrals numerically, but for this example, we will use the built-in function
Julia has for φ(x). It is in a package called Distributions:
In [2]: using Distributions
The following defines stdnormal as the standard normal random variable.
In [3]: stdnormal=Normal(0,1)
Out[3]: Normal{Float64}(μ=0.0, σ=1.0)
The built-in function cdf(stdnormal,x) computes the standard normal distribution
function at x. We write a function phi(x) based on this built-in function, matching
our notation φ(x) for the distribution function.
In [4]: phi(x)=cdf(stdnormal,x)
Out[4]: phi (generic function with 1 method)