Problem 2: Recall the formula for computing the price $C_0$ of an option (derivative of the BLM stock prices) That yields a payoff at time $T$, denote by $C_T$: $$ C_0 = \frac{1}{(1+r)^T}E^* (C_T), $$ where $*$ refers to the fact that we must use the value $p^*$ instead of the original (real) $P$ for the up/down probability of the BLM. (The real value of $P$ is not needed for priciing.) Also recall that for $C T = (S_T - K)^+$, the European call option, the expected value, $E^*(S_T-K)^+$ can be computed explicitly yielding the famous Black-Scholes-Merton option pricing formula: $$ C_0 = \frac{1}X(1+r)^T}\sum_{k=0}^{T} \left( \begin{array}{c} T \|| k lend{array} \right )(p^*)^k(1- p^*)^{T-k}(u^k d^{T-k}S_0-K)^+. $$ You are to use this formula to exactly obtain the price on the one hand, and then use Monte Carlo simulation on the other hand to compare and thus see how accurate the Monte Carlo method can be. Here are the parameters to use: $T = 10, r = 0.05, u = 1.15, d%3D 1.01, S_0 = 50, K = 70$. Recall that $$ p^* = \frac{1+r-d}{u-d}. $$ For the Monte Carlo, use $n=100, n=1000, n=10,000$ iid copies of $C_T$ (for averaging) to see how it gets more accurate as $n$ increases.

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Python Programming: Option Pricing Using Monte Carlo Simulation

Problem 2: Recall the formula for computing the price $C_0$ of an option (derivative of the BLM
stock prices) That yields a payoff at time $T$, denote by $C_T$: $$ C_0 = \frac{1}{(1+r)^T}E^*
%3D
(C_T), $$
where $*$ refers to the fact that we must use the value $p^*$ instead of the original (real) $P$ for
the up/down probability of the BLM. (The real value of $P$ is not needed for priciing.) Also recall
that for $C_T = (S_T - K)^+$, the European call option, the expected value, $E^*(S_T-K)^+$ can be
computed explicitly yielding the famous Black-Scholes-Merton option pricing formula:
$$ C_0 = \frac{1{(1+r)^T}\sum_{k=0}^{T} \left( \begin{array}{c} T \| k \end{array} \right )(p^*)^k(1-
p^*)^{T-k}(u^k d^{T-k}S_0-K)^+ . $$
You are to use this formula to exactly obtain the price on the one hand, and then use Monte Carlo
simulation on the other hand to compare and thus see how accurate the Monte Carlo method can
be.
Here are the parameters to use: $T = 10, r = 0.05, u = 1.15, d= 1.01, S_0 = 50, K = 70$. Recall that $$
%3D
p^* = \frac{1+r-d}{u-d} . $$
For the Monte Carlo, use $n=100, n=1000, n=10,000$ iid copies of $C_T$ (for averaging) to see how
it gets more accurate as $n$ increases.
In [3]:
# write your code here
# you may add more cells as needed
Transcribed Image Text:Problem 2: Recall the formula for computing the price $C_0$ of an option (derivative of the BLM stock prices) That yields a payoff at time $T$, denote by $C_T$: $$ C_0 = \frac{1}{(1+r)^T}E^* %3D (C_T), $$ where $*$ refers to the fact that we must use the value $p^*$ instead of the original (real) $P$ for the up/down probability of the BLM. (The real value of $P$ is not needed for priciing.) Also recall that for $C_T = (S_T - K)^+$, the European call option, the expected value, $E^*(S_T-K)^+$ can be computed explicitly yielding the famous Black-Scholes-Merton option pricing formula: $$ C_0 = \frac{1{(1+r)^T}\sum_{k=0}^{T} \left( \begin{array}{c} T \| k \end{array} \right )(p^*)^k(1- p^*)^{T-k}(u^k d^{T-k}S_0-K)^+ . $$ You are to use this formula to exactly obtain the price on the one hand, and then use Monte Carlo simulation on the other hand to compare and thus see how accurate the Monte Carlo method can be. Here are the parameters to use: $T = 10, r = 0.05, u = 1.15, d= 1.01, S_0 = 50, K = 70$. Recall that $$ %3D p^* = \frac{1+r-d}{u-d} . $$ For the Monte Carlo, use $n=100, n=1000, n=10,000$ iid copies of $C_T$ (for averaging) to see how it gets more accurate as $n$ increases. In [3]: # write your code here # you may add more cells as needed
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