## Exercises

The first few exercises concern pricing a European call option under the

assumption of risk neutrality. The price satisfies

$$

P = \beta^n \mathbb E \max\{ S_n - K, 0 \}

$$

where

1. $\beta$ is a discount factor,

2. $n$ is the expiry date,

2. $K$ is the strike price and

3. $\{S_t\}$ is the price of the underlying asset at each time $t$.

For example, if the call option is to buy stock in Amazon at strike price $K$, the owner has the right (but not the obligation) to buy 1 share in Amazon at price $K$ after $n$ days.

The payoff is therefore $\max\{S_n - K, 0\}$

The price is the expectation of the payoff, discounted to current value.

```{exercise-start}

:label: sp_ex01

```

Suppose that $S_n$ has the [log-normal](https://en.wikipedia.org/wiki/Log-normal_distribution) distribution with parameters $\mu$ and $\sigma$. Let $f$ denote the density of this distribution. Then

$$

P = \beta^n \int_0^\infty \max\{x - K, 0\} f(x) dx

$$

Plot the function

$$

g(x) = \beta^n \max\{x - K, 0\} f(x)

$$

over the interval $[0, 400]$ when `μ, σ, β, n, K = 4, 0.25, 0.99, 10, 40`.

```{hint}

:class: dropdown

From `scipy.stats` you can import `lognorm` and then use `lognorm(x, σ, scale=np.exp(μ)` to get the density $f$.

```

```{exercise-end}

```

```{solution-start} sp_ex01

:class: dropdown

```

Here's one possible solution

```{code-cell} ipython3

from scipy.integrate import quad

from scipy.stats import lognorm

μ, σ, β, n, K = 4, 0.25, 0.99, 10, 40

def g(x):

return β**n * np.maximum(x - K, 0) * lognorm.pdf(x, σ, scale=np.exp(μ))

x_grid = np.linspace(0, 400, 1000)

y_grid = g(x_grid)

fig, ax = plt.subplots()

ax.plot(x_grid, y_grid, label="$g$")

ax.legend()

plt.show()

```

```{solution-end}

```

```{exercise}

:label: sp_ex02

In order to get the option price, compute the integral of this function numerically using `quad` from `scipy.optimize`.

```

```{solution-start} sp_ex02

:class: dropdown

```

```{code-cell} ipython3

P, error = quad(g, 0, 1_000)

print(f"The numerical integration based option price is {P:.3f}")

```

```{solution-end}

```

```{exercise}

:label: sp_ex03

Try to get a similar result using Monte Carlo to compute the expectation term in the option price, rather than `quad`.

In particular, use the fact that if $S_n^1, \ldots, S_n^M$ are independent

draws from the lognormal distribution specified above, then, by the law of

large numbers,

$$ \mathbb E \max\{ S_n - K, 0 \}

\approx

\frac{1}{M} \sum_{m=1}^M \max \{S_n^m - K, 0 \}

$$

Set `M = 10_000_000`

```

```{solution-start} sp_ex03

:class: dropdown

```

Here is one solution:

```{code-cell} ipython3

M = 10_000_000

S = np.exp(μ + σ * np.random.randn(M))

return_draws = np.maximum(S - K, 0)

P = β**n * np.mean(return_draws)

print(f"The Monte Carlo option price is {P:3f}")

```

```{solution-end}

```

```{exercise}

:label: sp_ex1

In {ref}`this lecture <functions>`, we discussed the concept of {ref}`recursive function calls <recursive_functions>`.

Try to write a recursive implementation of homemade bisection function {ref}`described above <bisect_func>`.

Try to write a recursive implementation of the homemade bisection function {ref}`described above <bisect_func>`.

Test it on the function {eq}`root_f`.

```