Question

Claim amounts for wind damage to insured homes are independent random variables with common density f(x) = ( 3 x4 , x > 1 0 , otherwise where x is amount of claim in thousands. (a) find the probability that a claim is below average? [19/27] (b) suppose 3 claims will be made. what is the expected value of the largest of the three claims? [2.025] (c) suppose 3 claims will be made. what is the expected value of smallest of the three claims?[1.125]

Answer 1

`f_X(x)=\begin{cases}\frac3{x^4}&\text{for }x>1\\\\0&\text{otherwise}\end{cases}`

This distribution has expectation

`E[X]=\displaystyle\int_(-\infty)^\infty xf_X(x)\,\mathrm dx=\int_1^\infty\frac3{x^3}\,\mathrm dx=\frac32`

a. The probability that
`X` falls below the average/expectation is

`P\left(X<\frac32\right)=\displaystyle\int_(-\infty)^(3/2)f_X(x)\,\mathrm dx=\int_1^(3/2)\frac3{x^4}\,\mathrm dx=\boxed{(19)/(27)}`

b. Denote by
`X_((3))` the largest of the three claims
`X_1,X_2,X_3`. Then the density of this maximum order statistic is

`f_{X_((3))}(x)=3f_X(x)F_X(x)^2`

where
`F_X(x)=P(X\le x)` is the distribution function for
`X`. This is given by

`F_X(x)=\displaystyle\int_(-\infty)^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x<1\\\\\displaystyle\int_1^xf_X(t)\,\mathrm dt=1-\frac1{x^3}&\text{for }x\ge1\end{cases}`

So we have

`f_{X_((3))}(x)=\begin{cases}0&\text{for }x<1\\\frac9{x^4}\left(1-\frac1{x^3}\right)^2&\text{for }x>1\end{cases}`

and the expectation is

`E[X_((3))]=\displaystyle\int_(-\infty)^\infty xf_{X_((3))}(x)\,\mathrm dx=\int_1^\infty\frac9{x^3}\left(1-\frac1{x^3}\right)^2\,\mathrm dx=(81)/(40)=\boxed{2.025}`

c. Denote by
`X_((1))` the smallest of the three claims.
`X_((1))` has density

`f_{X_((1))}(x)=3f_X(x)(1-F_X(x))^2=\begin{cases}0&\text{for }x<1\\\\\frac9{x^(10)}&\text{for }x>1\end{cases}`

so the expectation is

`E[X_((1))]=\displaystyle\int_(-\infty)^\infty xf_{X_((1))}(x)\,\mathrm dx=\int_1^\infty\frac9{x^9}\,\mathrm dx=\frac98=\boxed{1.125}`