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Use vectorization and numba jitting for optimization #184

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Use vectorization and numba jitting for optimization #184

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Use vectorization and numba jitting for optimization
kp992 Nov 23, 2024
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remove ipywidgets because it takes too long to build and does not wor…
kp992 Nov 24, 2024
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80 changes: 40 additions & 40 deletions lectures/black_litterman.md
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Original file line number Diff line number Diff line change
Expand Up @@ -3,8 +3,10 @@ jupytext:
text_representation:
extension: .md
format_name: myst
format_version: 0.13
jupytext_version: 1.16.4
kernelspec:
display_name: Python 3
display_name: Python 3 (ipykernel)
language: python
name: python3
---
Expand Down Expand Up @@ -79,15 +81,13 @@ adjust investors' subjective beliefs about mean returns in order to render more

Let's start with some imports:

```{code-cell} ipython
```{code-cell} ipython3
import numpy as np
import scipy.stats as stat
import matplotlib.pyplot as plt
from ipywidgets import interact, FloatSlider
from numba import jit
```



## Mean-Variance Portfolio Choice

A risk-free security earns one-period net return $r_f$.
Expand Down Expand Up @@ -169,7 +169,7 @@ $w$'s with **extreme long and short positions**.
A common reaction to these outcomes is that they are so implausible that a portfolio
manager cannot recommend them to a customer.

```{code-cell} python3
```{code-cell} ipython3
np.random.seed(12)

N = 10 # Number of assets
Expand Down Expand Up @@ -300,7 +300,7 @@ The starting point of the Black-Litterman portfolio choice model is thus
a pair $(\delta_m, \mu_m)$ that tells the customer to hold the
market portfolio.

```{code-cell} python3
```{code-cell} ipython3
# Observed mean excess market return
r_m = w_m @ μ_est

Expand All @@ -316,11 +316,12 @@ d_m = r_m / σ_m
# Derive "view" which would induce the market portfolio
μ_m = (d_m * Σ_est @ w_m).reshape(N, 1)

x = np.arange(N) + 1
fig, ax = plt.subplots(figsize=(8, 5))
ax.set_title(r'Difference between $\hat{\mu}$ (estimate) and $\mu_{BL}$ (market implied)')
ax.plot(np.arange(N)+1, μ_est, 'o', c='k', label='$\hat{\mu}$')
ax.plot(np.arange(N)+1, μ_m, 'o', c='r', label='$\mu_{BL}$')
ax.vlines(np.arange(N) + 1, μ_m, μ_est, lw=1)
ax.plot(x, μ_est, 'o', c='k', label='$\hat{\mu}$')
ax.plot(x, μ_m, 'o', c='r', label='$\mu_{BL}$')
ax.vlines(x, μ_m, μ_est, lw=1)
ax.axhline(0, c='k', ls='--')
ax.set_xlabel('Assets')
ax.xaxis.set_ticks(np.arange(1, N+1, 1))
Expand Down Expand Up @@ -384,7 +385,7 @@ If $\hat \mu$ is the maximum likelihood estimator
and $\tau$ is chosen heavily to weight this view, then the
customer's portfolio will involve big short-long positions.

```{code-cell} python3
```{code-cell} ipython3
def black_litterman(λ, μ1, μ2, Σ1, Σ2):
"""
This function calculates the Black-Litterman mixture
Expand All @@ -402,10 +403,9 @@ def black_litterman(λ, μ1, μ2, Σ1, Σ2):

# The Black-Litterman recommendation for the portfolio weights
w_tilde = np.linalg.solve(δ * Σ_est, μ_tilde)
```

τ_slider = FloatSlider(min=0.05, max=10, step=0.5, value=τ)

@interact(τ=τ_slider)
```{code-cell} ipython3
def BL_plot(τ):
μ_tilde = black_litterman(1, μ_m, μ_est, Σ_est, τ * Σ_est)
w_tilde = np.linalg.solve(δ * Σ_est, μ_tilde)
Expand Down Expand Up @@ -439,6 +439,8 @@ def BL_plot(τ):
ax[1].xaxis.set_ticks(np.arange(1, N+1, 1))
ax[1].legend(numpoints=1)
plt.show()

BL_plot(τ)
```

## Bayesian Interpretation
Expand Down Expand Up @@ -607,7 +609,7 @@ $\bar d_2$ on the RHS of the constraint, by varying
$\bar d_2$ (or $\lambda$ ), we can trace out the whole curve
as the figure below illustrates.

```{code-cell} python3
```{code-cell} ipython3
np.random.seed(1987102)

N = 2 # Number of assets
Expand Down Expand Up @@ -641,21 +643,18 @@ r2 = np.linspace(-0.02, .15, N_r2)
curve = np.asarray([black_litterman(λ, μ_m, μ_est, Σ_est,
τ * Σ_est).flatten() for λ in λ_grid])

λ_slider = FloatSlider(min=.1, max=7, step=.5, value=1)
λ = 1
```

@interact(λ=λ_slider)
```{code-cell} ipython3
def decolletage(λ):
dist_r_BL = stat.multivariate_normal(μ_m.squeeze(), Σ_est)
dist_r_hat = stat.multivariate_normal(μ_est.squeeze(), τ * Σ_est)

X, Y = np.meshgrid(r1, r2)
Z_BL = np.zeros((N_r1, N_r2))
Z_hat = np.zeros((N_r1, N_r2))

for i in range(N_r1):
for j in range(N_r2):
Z_BL[i, j] = dist_r_BL.pdf(np.hstack([X[i, j], Y[i, j]]))
Z_hat[i, j] = dist_r_hat.pdf(np.hstack([X[i, j], Y[i, j]]))
XY = np.stack((X, Y), axis=-1)
Z_BL = dist_r_BL.pdf(XY)
Z_hat = dist_r_hat.pdf(XY)

μ_tilde = black_litterman(λ, μ_m, μ_est, Σ_est, τ * Σ_est).flatten()

Expand All @@ -676,6 +675,8 @@ def decolletage(λ):
ax.text(μ_est[0] + 0.003, μ_est[1], r'$\hat{\mu}$')
ax.text(μ_m[0] + 0.003, μ_m[1] + 0.005, r'$\mu_{BL}$')
plt.show()

decolletage(λ)
```

Note that the line that connects the two points
Expand All @@ -692,26 +693,22 @@ This leads to the
following figure, on which the curve connecting $\hat \mu$
and $\mu_{BL}$ are bending

```{code-cell} python3
```{code-cell} ipython3
λ_grid = np.linspace(.001, 20000, 1000)
curve = np.asarray([black_litterman(λ, μ_m, μ_est, Σ_est,
τ * np.eye(N)).flatten() for λ in λ_grid])
λ = 200
```

λ_slider = FloatSlider(min=5, max=1500, step=100, value=200)

@interact(λ=λ_slider)
```{code-cell} ipython3
def decolletage(λ):
dist_r_BL = stat.multivariate_normal(μ_m.squeeze(), Σ_est)
dist_r_hat = stat.multivariate_normal(μ_est.squeeze(), τ * np.eye(N))

X, Y = np.meshgrid(r1, r2)
Z_BL = np.zeros((N_r1, N_r2))
Z_hat = np.zeros((N_r1, N_r2))

for i in range(N_r1):
for j in range(N_r2):
Z_BL[i, j] = dist_r_BL.pdf(np.hstack([X[i, j], Y[i, j]]))
Z_hat[i, j] = dist_r_hat.pdf(np.hstack([X[i, j], Y[i, j]]))
XY = np.stack((X, Y), axis=-1)
Z_BL = dist_r_BL.pdf(XY)
Z_hat = dist_r_hat.pdf(XY)

μ_tilde = black_litterman(λ, μ_m, μ_est, Σ_est, τ * np.eye(N)).flatten()

Expand All @@ -733,6 +730,8 @@ def decolletage(λ):
ax.text(μ_est[0] + 0.003, μ_est[1], r'$\hat{\mu}$')
ax.text(μ_m[0] + 0.003, μ_m[1] + 0.005, r'$\mu_{BL}$')
plt.show()

decolletage(λ)
```

## Black-Litterman Recommendation as Regularization
Expand Down Expand Up @@ -1247,7 +1246,7 @@ observations is related to the sampling frequency

- Moreover, for a fixed lag length, $n$, the dependence vanishes as the sampling frequency goes to infinity. In fact, letting $h$ go to $\infty$ gives back the case of IID data.

```{code-cell} python3
```{code-cell} ipython3
μ = .0
κ = .1
σ = .5
Expand Down Expand Up @@ -1346,7 +1345,8 @@ thus getting an idea about how the asymptotic relative MSEs changes in
the sampling frequency $h$ relative to the IID case that we
compute in closed form.

```{code-cell} python3
```{code-cell} ipython3
@jit
def sample_generator(h, N, M):
ϕ = (1 - np.exp(-κ * h)) * μ
ρ = np.exp(-κ * h)
Expand All @@ -1355,18 +1355,18 @@ def sample_generator(h, N, M):
mean_uncond = μ
std_uncond = np.sqrt(σ**2 / (2 * κ))

ε_path = stat.norm(0, np.sqrt(s)).rvs((M, N))
ε_path = np.random.normal(0, np.sqrt(s), (M, N))

y_path = np.zeros((M, N + 1))
y_path[:, 0] = stat.norm(mean_uncond, std_uncond).rvs(M)
y_path[:, 0] = np.random.normal(mean_uncond, std_uncond, M)

for i in range(N):
y_path[:, i + 1] = ϕ + ρ * y_path[:, i] + ε_path[:, i]

return y_path
```

```{code-cell} python3
```{code-cell} ipython3
# Generate large sample for different frequencies
N_app, M_app = 1000, 30000 # Sample size, number of simulations
h_grid = np.linspace(.1, 80, 30)
Expand Down
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