Order statistics, what am I doing wrong

Question

From SOA sample 138:

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

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

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 $Y$:

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

Then $P(X \le k)$ is

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

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

$$\left({10k\over50}-{k^2\over100}\right)^2$$

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

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

$${200k\over2500}-{60k^2\over5000}+{4k^3\over10000}$$

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

$$E(K)=\int_0^{10} k\left({200k\over2500}-{60k^2\over5000}+{4k^3\over10000}\right)dk$$

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

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

$$=4.666$$

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

Answer 1

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 $P(X \leq k)P(Y \leq k) = \frac9{16}.$

Answer 2

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