Abou independent random variables

Question

Problem:Suppose two independent claims are made on two insured homes, where each claim has PDF $$f(x)=\begin{cases} \frac{4}{x^5}, & 1<x<\infty \\ 0, & \text{elsewhere} \end{cases}$$in which the unit is 1000. Find the expected value of the larger claim.

Thoughts:We are given PDF, and so we need to find CDF of y. Since $X_1$ and $X_2$ are independent claims and we can let Y=max { $X_1,X_2$}, how am I going to find the CDF? Do I need to find by cases, like I find for $1<x< \infty$ to be $1-\frac{x}{x^4}$ and 0 to be any constant C. I am not sure the general idea of approaching for this problem.

Answer 1

If you are not familiar with conditional expectation we can proceed as follows. First we find the cdf of $X_1$. For $x> 1$ $$ P(X_1\leq x)=\int_1^x4t^{-5}\, dt=1-\frac{1}{x^4}. $$ Let $Y=\max(X_1, X_2)$. Then $Y$ has cdf given by $$ \begin{align} P(Y\leq y)&=P(\max(X_1, X_2)\leq y)\\ &=P(X_1\leq y, X_2\leq y)\\ &=P(X_1\leq y)^2\tag{0}\\ &=\left(1-y^{-4}\right)^2\tag{1} \end{align} $$

for $y>1$ where we used the independent and identically distributed assumption in (0). At this point you can find the density of $Y$, say $f$, by differentiating $(1)$ and compute $$ EY=\int_1^\infty yf(y)\, dy $$ which I leave to you.

Answer 2

You do not need to find the CDF of $Y$.

Just use linearity of expectation, and symmetry (as the variables as independent and identically distributed).

$$\begin{align}\mathsf E(\max\{X_1,X_2\}) &=\mathsf E(X_1\mathbf 1_{X_1>X_2}+X_2\mathbf 1_{X_1\leqslant X_2})\\[2ex]&=\mathsf E(X_1\mathbf 1_{X_1>X_2})+\mathsf E(X_2\mathbf 1_{X_1\leqslant X_2})\\[2ex]&=2~\mathsf E(X_1\mathbf 1_{X_1>X_2})\\[2ex]&=2~\iint_{s>t} s\, f(s) f(t)~\mathrm d (s,t)\\[2ex]&=2\int_1^\infty\int_1^s \dfrac{16}{s^4\,t^5}~\mathrm d t~\mathrm d s\\[2ex]&~~\vdots\end{align}$$