Question

An insurance company sells automobile liability and collision insurance. Let X denote the percentage of liability policies that will be renewed at the end of their terms and Y the percentage of collision policies that will be renewed at the end of their terms. X and Y have the joint cumulative distribution function F(x,y) = \dfrac{xy(x+y)}{2{,}000{,}000}, 0\le x\le 100, 0\le y \le 100. Calculate Var(X).

Answer 1

Answer:

-764.28

Step-by-step explanation:

Given the joint cumulative distribution of X and Y as

$F(x,y) = \frac{xy(x+y)}{2000000}\ \ \ \, \ 0\leq x100, 0\leq y\leq 100$

#First find $F_x$ and probability distribution function ,$f_x(x)$:

$F_x(x)=F(x,100)\\\\\\=\frac{100x(x+100)}{2000000}\\\\\\\\=\frac{100x^2+10000x}{2000000}\\\\\\=>f_x(x)=\frac{x}{10000}+\frac{1}{200}$

#Have determined the probability distribution unction ,$f_x(x)$, we calculate the Expectation of the random variable X:

$E(X)=\int\limits^{100}_0 \frac{x^2}{10000}+\frac{x}{200} dx \\\\\\\\=|\frac{x^3}{30000}+\frac{x^2}{400}|\limits^{100}_0\\\\=58.33$

#We then calculate $E(X^2)$:

$E(X^2)=\int\limits^{100}_0 \frac{x^3}{10000}+\frac{x^2}{200}\ dx\\\\=\frac{x^4}{40000}+\frac{x^3}{600}|\limits^{100}_0=4166.67\\\\Var(X)=E(X^2)-(E(X))^2=4166.67-58.33^2\\\\Var(X)=764.28$

Hence, the Var(X) is 764.28