Question

Monte Carlo Simulation This lab assignment consists of two problems involving Monte Carlo simulation. You should use either Excel or Python for this assignment. References: Seila, A. F., V. Ceric, and P. Tadikamalla, Applied Simulation Modeling, Duxbury - Brooks/Cole, Belmont, CA, 2003. Schriber, T. J., "Simulation for the Masses: Spreadsheet-based Monte Carlo Simulation," Proceedings of the 2009 Winter Simulation Conference, Rossetti, Hill, Johansson, Dunkin, and Ingalls, Eds., December 2009. Adopted from Problem 2.13 from Seila et al. (page 74) with minor changes 1. The actuaries for an insurance company represent the total losses experienced during a week as follows: Let N represent the number of policies that experience a loss during the week, and Z1, Z2, ... be independent random variables that have the same distribution representing the amount of each loss. Then the total losses for the week are X Z1 +Z2Z3 + ... + ZN --70CO maciCC ccvar c) Using probability theory, one can show that if N and Zı, Z2, ... are statistically independent, the expected value of X is E(X) E(N)*E(Z) = Compare the confidence interval of the mean from part b to the value computed from the expression for E(X) d) Estimate the probability that the loss will be less than $40,000 and the probability that the loss will be greater than $70,000.

Answer 1

a)The Python code for doing the Monte Carlo Simulation and finding the sample mean, sample standard deviation, 25th and 75th percentiles, 95% confidence interval and the probabilities is given below.

import math
import numpy as np
X = []
N = np.random.randint(low=50, high=145, size=5000)
for i in range(5000):
    s = 0
    for j in range(N[i]):
        Z = np.random.triangular(left=280, right=740, mode=510, size=1)
        s += Z
    X.append(s)
m= np.mean(X)
sd = np.std(X)
print("Sample mean is:", m)
print("Sample standard deviation is:", sd)
conf1 = m - 1.96*sd/np.sqrt(5000)
conf2 = m + 1.96*sd/np.sqrt(5000)
print("The 95% confidence interval is:", conf1, " ,", conf2)
print("The 25th and 75th percentiles are:", np.percentile(X, 25), " ,", np.percentile(X, 75))
X = np.array(X)
X1 = X[X < 40000]
X2 = X[X > 70000]
print("The probability that the loss will be less than $40,000 is:", len(X1)/5000)
print("The probability that the loss will be greater than $70,000 is:", len(X2)/5000)

The sample output is:

Sample mean is: 49859.333663659585
Sample standard deviation is: 13889.666862680726
The 95% confidence interval is: 49474.33174068026 , 50244.33558663891
The 25th and 75th percentiles are: 38091.16156246797 , 61866.255521021754
The probability that the loss will be less than $40,000 is: 0.2872
The probability that the loss will be greater than $70,000 is: 0.0776

b) We have sample mean is
X $49,859.334
, sample standard deviation is
S $13,889.667
, 95% confidence interval is
(49,474.332,50,244.336)
, 25th and 75th percentiles are
38,091.162, 61,866.256
, and the probabilities are
P(X 40,000) = 0.2872, P(X> 70,000) = 0.0776

c) We have

E(X) E(N)E(Z) E(X) =50+ 145 2 280 510 740 3 E(X) 49725

The 95% CI is

E(X2) E(N2)E(Z2) 50145 2802 5102 7402 280(510) + 280(740) + 740(510) 952 280510 740 E(X2) 18 12 E(X2) Var(X) 2758636806- 497252 Var(X) 286061 1818 = 2758636806 g16913.343

Xt 1.96- Vn 49725 1,9616913.343 V5000 (49, 256.19, 50,193.81)

The CI has shifted to left.

d)The probabilities are

P(X 40,000) = 0.2872, P(X> 70,000) = 0.0776