Cfrgtkky

From SOA sample 138:

A machine consists of two components, whose lifetimes have the joint density function

$$f(x,y)=
begin{cases}
{1over50}, & text{for }x>0,y>0,x+y<10 \
0, & text{otherwise}
end{cases}$$

The machine operates until both components fail.
Calculate the expected operational time of the machine.

I know there is a simpler way to solve this, but I would like to solve it using order statistics. This is what I have so far:

If I am understanding correctly, I need to find probability $P(max(X,Y) le k)=P(X le k)P(Y le k)$, and then differentiate to find the density of $k$, and from there find the expected value of $k$.

So first I will find the marginal density of $X$ and $Y$:

$$f(x) = int_0^{10-x}{1over50}dy$$
$$f(x) = {{10-x}over50} $$

Then $P(X le k)$ is

$$P(X le k) = int_0^{k}{{10-x}over50}dx $$
$$ = left({10xover50}-{x^2over100}right)bigg|_0^k$$
$$ = left({10kover50}-{k^2over100}right)$$

I can do the same thing for $y$, so $P(X le k)P(Y le k)$ is

$$left({10kover50}-{k^2over100}right)^2$$

$$={100k^2over2500}-{20k^3over5000}+{k^4over10000}$$

I will now take the derivative to get $f(k)$

$${200kover2500}-{60k^2over5000}+{4k^3over10000}$$

And now I can integrate to the limit of $k$ to get $E(K)$

$$E(K)=int_0^{10} kleft({200kover2500}-{60k^2over5000}+{4k^3over10000}right)dk$$

$$=int_0^{10} {200k^2over2500}-{60k^3over5000}+{4k^4over10000}dk$$

$$=left( {200k^3over7500}-{60k^4over20000}+{4k^5over50000}right)bigg|_0^{10}$$

$$=4.666$$

However the true solution is $5$, where did I go wrong?

The equation
$$P(max(X,Y) leq k)=P(X leq k)P(Y leq k)$$
would be true if $X$ and $Y$ were independent.
But they clearly are not independent.

There are multiple ways to confirm the dependence between $X$ and $Y$, but perhaps the most obvious is to choose a not-too-large value of $k$, compute both sides of the equation, and compare the results.

For example, with $k=5$ we get $P(max(X,Y) leq k) = frac12$
but $P(X leq k) = P(Y leq k) = frac34$ and therefore
$P(X leq k)P(Y leq k) = frac9{16}.$

David K has explained the fallacy, but I'll derive the answer you wanted. Let $S$ denote the $fne 0$ region; let $S'$ denote the subset of $S$ with $xle y$. Then$$frac{1}{50}int_Smax{x,,y}dxdy=frac{1}{25}int_{S'}ydxdy=frac{1}{25}int_0^{10}ymin{y,,10-y}dy\=frac{1}{25}left(int_0^5y^2dy+int_5^{10}(10y-y^2)dyright)=5.$$