Question

Please solve step by step. USE PYTHON OR EXCELL NO OTHERS.

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 Zı, Z2, independent random variables that have the same distribution representing the amount of each loss be Then the total losses for the week are Z1+Z2+Z3 ... + ZN X a) Let N have a discrete uniform distribution over the values 50 to 145 and let each Z, have a symmetric triangular distribution between $280 and $740. Develop a Monte Carlo simulation to estimate the total weekly loss and run the simulation for 5000 independent replications. b) Using the data from the simulation, compute the sample mean, sample standard deviation, the 95% confidence interval on the mean and the 25th and 75th percentiles of the sample. c) Using probability theory, one can show that if N and Zi, 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 $\fn_jvn {\color{Blue} \overline{X}=\$49,859.334}$ , sample standard deviation is $\fn_jvn {\color{Blue} s=\$13,889.667 }$ , 95% confidence interval is $\fn_jvn {\color{Blue} \left ( 49,474.332 , 50,244.336 \right ) }$ , 25th and 75th percentiles are $\fn_jvn {\color{Blue} 38,091.162 , 61,866.256}$ , and the probabilities are $\fn_jvn {\color{Blue} P(X>40,000)= 0.2872,P(X>70,000)=0.0776}$

c) We have

$\fn_jvn E(X)=E(N)E(Z)\\ E(X)=\left ( \frac{50+145}{2} \right )\left ( \frac{280+510+740}{3} \right )\\ E(X)=49 725$

$\fn_jvn E(X^2)=E(N^2)E(Z^2)\\ E(X^2)=\left [\left ( \frac{50+145}{2} \right )^2 +\frac{95^2}{12} \right ]\left [\left ( \frac{280+510+740}{3} \right )^2+\frac{280^2+510^2+740^2-280(510)+280(740)+740(510)}{18} \right ]\\ E(X^2)=2 758 636 806\\ Var(X)=2 758 636 806-49725^2\\ Var(X)=286 061 1818\\ \sigma =16 913.343$The 95% CI is

$\fn_jvn \overline{X}\pm 1.96\frac{\sigma }{\sqrt{n}}\\ 49725 \pm 1.96\frac{16 913.343}{\sqrt{5000}}\\ {\color{Blue} \left ( 49,256.19, 50,193.81 \right )}$

The CI has shifted to left.

d)The probabilities are $\fn_jvn {\color{Blue} P(X>40,000)= 0.2872,P(X>70,000)=0.0776}$