Update rng usage in numba.md

Co-Authored-By: Claude Sonnet 4.6 <noreply@anthropic.com>

Author: Chihiro2000GitHub

lectures/numba.md

@@ -454,11 +454,13 @@ Compare speed with and without Numba when the sample size is large.
 Here is one solution:
 
 ```{code-cell} ipython3
+rng = np.random.default_rng()
+
 @jit
-def calculate_pi(n=1_000_000):
+def calculate_pi(rng, n=1_000_000):
     count = 0
     for i in range(n):
-        u, v = np.random.uniform(0, 1), np.random.uniform(0, 1)
+        u, v = rng.uniform(0, 1), rng.uniform(0, 1)
         d = np.sqrt((u - 0.5)**2 + (v - 0.5)**2)
         if d < 0.5:
             count += 1
@@ -471,12 +473,12 @@ Now let's see how fast it runs:
 
 ```{code-cell} ipython3
 with qe.Timer():
-    calculate_pi()
+    calculate_pi(rng)
 ```
 
 ```{code-cell} ipython3
 with qe.Timer():
-    calculate_pi()
+    calculate_pi(rng)
 ```
 
 If we switch off JIT compilation by removing `@jit`, the code takes around
@@ -550,10 +552,12 @@ p, q = 0.1, 0.2  # Prob of leaving low and high state respectively
 Here's a pure Python version of the function
 
 ```{code-cell} ipython3
-def compute_series(n):
+rng = np.random.default_rng()
+
+def compute_series(n, rng):
     x = np.empty(n, dtype=np.int64)
     x[0] = 1  # Start in state 1
-    U = np.random.uniform(0, 1, size=n)
+    U = rng.uniform(0, 1, size=n)
     for t in range(1, n):
         current_x = x[t-1]
         if current_x == 0:
@@ -568,7 +572,7 @@ state is about 0.666
 
 ```{code-cell} ipython3
 n = 1_000_000
-x = compute_series(n)
+x = compute_series(n, rng)
 print(np.mean(x == 0))  # Fraction of time x is in state 0
 ```
 
@@ -578,7 +582,7 @@ Now let's time it:
 
 ```{code-cell} ipython3
 with qe.Timer():
-    compute_series(n)
+    compute_series(n, rng)
 ```
 
 Next let's implement a Numba version, which is easy
@@ -590,15 +594,15 @@ compute_series_numba = jit(compute_series)
 Let's check we still get the right numbers
 
 ```{code-cell} ipython3
-x = compute_series_numba(n)
+x = compute_series_numba(n, rng)
 print(np.mean(x == 0))
 ```
 
 Let's see the time
 
 ```{code-cell} ipython3
 with qe.Timer():
-    compute_series_numba(n)
+    compute_series_numba(n, rng)
 ```
 
 This is a nice speed improvement for one line of code!
@@ -636,11 +640,17 @@ For the size of the Monte Carlo simulation, use something substantial, such as
 Here is one solution:
 
 ```{code-cell} ipython3
+n = 1_000_000
+rng = np.random.default_rng()
+u_draws = rng.uniform(size=n)
+v_draws = rng.uniform(size=n)
+
 @jit(parallel=True)
-def calculate_pi(n=1_000_000):
+def calculate_pi(u_draws, v_draws):
+    n = len(u_draws)
     count = 0
     for i in prange(n):
-        u, v = np.random.uniform(0, 1), np.random.uniform(0, 1)
+        u, v = u_draws[i], v_draws[i]
         d = np.sqrt((u - 0.5)**2 + (v - 0.5)**2)
         if d < 0.5:
             count += 1
@@ -653,12 +663,12 @@ Now let's see how fast it runs:
 
 ```{code-cell} ipython3
 with qe.Timer():
-    calculate_pi()
+    calculate_pi(u_draws, v_draws)
 ```
 
 ```{code-cell} ipython3
 with qe.Timer():
-    calculate_pi()
+    calculate_pi(u_draws, v_draws)
 ```
 
 By switching parallelization on and off (selecting `True` or
@@ -773,6 +783,7 @@ def compute_call_price_parallel(β=β,
         s = np.log(S0)
         h = h0
         # Simulate forward in time
+        # Draws are kept inside the loop to avoid pre-allocating large shock arrays.
         for t in range(n):
             s = s + μ + np.exp(h) * np.random.randn()
             h = ρ * h + ν * np.random.randn()