Minimising and root finding

Two jobs: find the lowest point, find where a curve crosses zero

scipy.optimize solves two kinds of problems. The first is minimising a function $f(x)$ , that is, finding the input with the smallest output. You call minimize(f, x0) where x0 is your starting guess. The method BFGS follows the slope of $f$ downhill. Nelder-Mead needs no slope and just compares values. SLSQP also accepts bounds and constraints, such as weights that must add up to 1. The answer is only a local minimum, the bottom of the valley where you started, and a different x0 can give a different answer. Always check res.success.

The second problem is root finding: solving $g(x)=0$ . brentq(g, a, b) needs a bracket, two points where $g$ has opposite signs, and then it is sure to find a root by repeatedly halving the interval. After $n$ halvings the interval is $(b-a)/2^n$ wide. newton(g, x0, fprime) uses the slope of $g$ and is much faster, but it needs a good start and can jump far away when the slope is nearly flat. In finance the yield to maturity of a bond is a root: it is the rate $y$ that makes the bond's price formula equal to its market price. A Newton step is $y_{\text{new}}=y-g(y)/g'(y)$ .

y_{\text{new}}=y-\frac{g(y)}{g'(y)},\qquad g(y)=\frac{5}{1+y}+\frac{105}{(1+y)^{2}}-98,\qquad \text{width after }n\text{ halvings}=\frac{b-a}{2^{n}}

(1)

The yield of a two-year bondA bond pays

5

after one year and

105

after two years, and it costs

98

. Start at

y=6\%

. The price formula gives

98.17

, so

g=0.17

. The slope is

g'=-5/1.06^{2}-210/1.06^{3}=-180.8

. One Newton step gives

y=0.06-0.17/(-180.8)=6.092\%

, and brentq on the bracket

[0,0.5]

confirms the root

6.0923\%

. For bisection on the bracket

[0,1]

to be within

0.001

you need

2^{n}\ge1000

, so

n=10

halvings. For a minimum, two assets with volatilities

20

and

10

and correlation

0.2

have a covariance of

40

, so the minimum-variance weight in the first is

(100-40)/(400+100-80)=0.143

. On real data, SPY and TLT over

5{,}404

days from February 2005 have volatilities of

19.0\%

and

14.6\%

a year and a correlation of

-0.30

. The minimum-variance weight in SPY is

40\%

, giving a volatility of

9.7\%

. For a function with two valleys, such as

x^{4}-3x^{2}+x

, minimize from

x_0=2

stops at

1.13

and from

x_0=-2

-1.30

, which is the lower one.

Minimise or solve: minimize finds a lowest point, brentq and newton find where a function is zero.
A bracket is a guarantee: brentq always converges when the signs differ, while newton is faster but needs a good start.
A minimum can be local: try several starting points and read res.success before trusting res.x.

Lab · what's graded

One Newton step15 pts
Bisection halvings10 pts
A minimum-variance weight15 pts
The right tool for each job15 pts
An implied-volatility workflow10 pts
A possibly local minimum10 pts
Why a bracket, and why Newton can fail25 pts

Two jobs: find the lowest point, find where a curve crosses zero

y_{\text{new}}=y-\frac{g(y)}{g'(y)},\qquad g(y)=\frac{5}{1+y}+\frac{105}{(1+y)^{2}}-98,\qquad \text{width after }n\text{ halvings}=\frac{b-a}{2^{n}}

(1)

The yield of a two-year bondA bond pays

5

after one year and

105

after two years, and it costs

98

. Start at

y=6\%

. The price formula gives

98.17

, so

g=0.17

. The slope is

g'=-5/1.06^{2}-210/1.06^{3}=-180.8

. One Newton step gives

y=0.06-0.17/(-180.8)=6.092\%

, and brentq on the bracket

[0,0.5]

confirms the root

6.0923\%

. For bisection on the bracket

[0,1]

to be within

0.001

you need

2^{n}\ge1000

, so

n=10

halvings. For a minimum, two assets with volatilities

20

and

10

and correlation

0.2

have a covariance of

40

, so the minimum-variance weight in the first is

(100-40)/(400+100-80)=0.143

. On real data, SPY and TLT over

5{,}404

days from February 2005 have volatilities of

19.0\%

and

14.6\%

a year and a correlation of

-0.30

. The minimum-variance weight in SPY is

40\%

, giving a volatility of

9.7\%

. For a function with two valleys, such as

x^{4}-3x^{2}+x

, minimize from

x_0=2

stops at

1.13

and from

x_0=-2

-1.30

, which is the lower one.

Minimise or solve: minimize finds a lowest point, brentq and newton find where a function is zero.
A bracket is a guarantee: brentq always converges when the signs differ, while newton is faster but needs a good start.
A minimum can be local: try several starting points and read res.success before trusting res.x.

Lab · what's graded

One Newton step15 pts
Bisection halvings10 pts
A minimum-variance weight15 pts
The right tool for each job15 pts
An implied-volatility workflow10 pts
A possibly local minimum10 pts
Why a bracket, and why Newton can fail25 pts

Two jobs: find the lowest point, find where a curve crosses zero

Lab · what's graded

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Minimising and root finding

Two jobs: find the lowest point, find where a curve crosses zero

Lab · what's graded

Log in to run your code.