suggestion

Author: Smit-create

lectures/monte_carlo.md

@@ -12,7 +12,6 @@ kernelspec:
 ---
 
 
-
 # Monte Carlo and Option Pricing
 
 ## Overview
@@ -49,7 +48,6 @@ from numpy.random import randn
 ```
 
 
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 ## An Introduction to Monte Carlo
 
 In this section we describe how Monte Carlo can be used to compute
@@ -153,7 +151,6 @@ p = 0.5
 ```
 
 
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 #### A Routine using Loops in Python
 
 
@@ -177,7 +174,6 @@ S / n
 ```
 
 
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 We can also construct a function that contains these operations:
 
 ```{code-cell} ipython3
@@ -192,15 +188,13 @@ def compute_mean(n=1_000_000):
 ```
 
 
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 Now let's call it.
 
 ```{code-cell} ipython3
 compute_mean()
 ```
 
 
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 ### A Vectorized Routine
 
 If we want a more accurate estimate we should increase $n$.
@@ -225,7 +219,6 @@ compute_mean_vectorized()
 ```
 
 
-
 Notice that this routine is much faster.
 
 We can increase $n$ to get more accuracy and still have reasonable speed:
@@ -237,7 +230,6 @@ compute_mean_vectorized(n=10_000_000)
 ```
 
 
-
 ## Pricing a European Call Option under Risk Neutrality
 
 Next we are going to price a European call option under risk neutrality.
@@ -284,7 +276,6 @@ $$
 $$
 
 
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 ### A Comment on Risk
 
 As suggested by the name, the risk-neutral price ignores risk.
@@ -306,7 +297,6 @@ Nonetheless, the risk-neutral price is an important benchmark, which economists
 and financial market participants try to calculate every day.
 
 
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 ### Discounting
 
 Another thing we ignored in the previous discussion was time.
@@ -336,7 +326,6 @@ $$
 $$
 
 
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 ### European Call Options
 
 Now let's price a European call option.
@@ -388,15 +377,13 @@ n = 10
 ```
 
 
-
 We set the simulation size to
 
 ```{code-cell} ipython3
 M = 10_000_000
 ```
 
 
-
 Here is our code
 
 ```{code-cell} ipython3
@@ -407,7 +394,6 @@ print(f"The Monte Carlo option price is approximately {P:3f}")
 ```
 
 
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 ## Pricing Via a Dynamic Model
 
 In this exercise we investigate a more realistic model for the share price $S_n$.
@@ -479,7 +465,6 @@ $$
 Here $\{\eta_t\}$ is also IID and standard normal.
 
 
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 ### Default Parameters
 
 For the dynamic model, we adopt the following parameter values.
@@ -493,7 +478,6 @@ h0 = 0
 ```
 
 
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 (Here `S0` is $S_0$ and `h0` is $h_0$.)
 
 For the option we use the following defaults.
@@ -505,7 +489,6 @@ n = 10
 ```
 
 
-
 ### Visualizations
 
 With $s_t := \ln S_t$, the price dynamics become
@@ -528,7 +511,6 @@ def simulate_asset_price_path(μ=μ, S0=S0, h0=h0, n=n, ρ=ρ, ν=ν):
 ```
 
 
-
 Here we plot the paths and the log of the paths.
 
 ```{code-cell} ipython3
@@ -547,7 +529,6 @@ plt.show()
 ```
 
 
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 ### Computing the Price
 
 Now that our model is more complicated, we cannot easily determine the
@@ -598,7 +579,6 @@ compute_call_price()
 ```
 
 
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 ## Exercises
 
 ```{exercise}
@@ -644,7 +624,6 @@ compute_call_price()
 ```
 
 
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 Notice that this version is faster than the one using a Python loop.
 
 Now let's try with larger $M$ to get a more accurate calculation.
@@ -655,7 +634,6 @@ compute_call_price(M=10_000_000)
 ```
 
 
-
 ```{solution-end}
 ```
 
@@ -703,17 +681,18 @@ def compute_call_price_with_barrier(β=β,
     for m in range(M):
         s = np.log(S0)
         h = h0
-        is_null = False
+        payoff = 0
         # Simulate forward in time
         for t in range(n):
             s = s + μ + np.exp(h) * randn()
             h = ρ * h + ν * randn()
             if np.exp(s) > bp:
-                is_null = True
-
-        if not is_null:
-            # And add the value max{S_n - K, 0} to current_sum
-            current_sum += np.maximum(np.exp(s) - K, 0)
+                payoff = 0
+                break
+            else:
+                payoff = np.maximum(np.exp(s) - K, 0)
+        # And add the payoff to current_sum
+        current_sum += payoff
 
     return β**n * current_sum / M
 ```