Question

An insurance company offers two types of insurance: health insurance and life insurance. Let X be the annual claim on the health insurance and Y be the annual claim on the life insurance. The joint density function of X and Y is: Find the probability that the total claims exceed 0.5 but is less than 2 and the claim on the life insurance is less than 1.5.

Answer 1

The joint density equation is missing in the question. The equation is $f(x,y)=\left\{\begin{matrix}2x^3, 0\leq x\leq 1,0\leq y\leq 2\\ 0, \text{ otherwise}\end{matrix}\right.$

Probability is 0.59

Step-by-step explanation:

We have

x -- denotes the annual claim health insurance

y -- denotes the annual claim life insurance

And joint density of x and y is given to us

$f(x,y)=\left\{\begin{matrix}2x^3, 0\leq x\leq 1,0\leq y\leq 2\\ 0, \text{ otherwise}\end{matrix}\right.$

Event that the claim exceeds 0.5 but less than 2 can be written as 0.5≤ x+y ≤2

and claim of life insurance less than 1.5 can be written as 0 ≤ y ≤ 1.5

Required probability

= P( 0.5≤ x+y ≤2  ∩ 0 ≤ y ≤ 1.5 )

Lets plot this, to find the common region

For common region

when 0≤ x ≤ 0.5, then 0.5 ≤ y ≤ 1.5  ---- for region I

when 0.5≤ x ≤ 1, then 0 ≤ y ≤ 2-x  ---- for region II

The required probability

= $\int_{0}^{0.5} \int_{0.5-x}^{1.5}2x^3 dy dx+\int_{0.5}^{1} \int_{0}^{2-x}2x^3 dy dx$

= $\int_{0}^{0.5} 2x^3[\int_{0.5-x}^{1.5}dy] dx+\int_{0.5}^{1} 2x^3[\int_{0}^{2-x}dy] dx$

= $\int_{0}^{0.5} 2x^3(1.5-0.5+x) dx+\int_{0.5}^{1} 2x^3 (2-x) dx$

= $\int_{0}^{0.5}2x^3 dx+\int_{0}^{0.5}2x^4 dx+\int_{0.5}^{1}4x^3 dx-\int_{0.5}^{1}2x^4$

= $[\frac{2x^4}{4}]_0^{0.5}+[\frac{2x^5}{5}]_0^{0.5}+[\frac{4x^4}{4}]_{0.5}^1-[\frac{2x^5}{5}]_{0.5}^1$

= $\frac{1}{32}+\frac{1}{80}+\frac{1}{16}-\frac{31}{20}$

= 19/32

=0.59