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

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`

=
`[(2x^4)/(4)]_0^(0.5)+[(2x^5)/(5)]_0^(0.5)+[(4x^4)/(4)]_(0.5)^1-[(2x^5)/(5)]_(0.5)^1`

=
`$(1)/(32)+(1)/(80)+(1)/(16)-(31)/(20)$`

= 19/32

=0.59