Calculate expected value of random variable $Z=\min\left\{X,Y\right\}$ Hi math people in my last question I  asked the question

$X,Y$ are independent, identical distributed with $$P(X=k) =
P(Y=k)=\frac{1}{2^k} \,\,\,\,\,\,\,\,\,\,\,\, (k=1,2,...,n,...)$$
Calculate the probabilities $P(\min\left\{X,Y\right\} \leq x)$

I calculate it correct, here is calculation:
$$\begin{split}F_M(x) &= P(\min\left\{X,Y\right\} \leq x) \\ &= 1-P(x<\min\left\{X,Y\right\} ) \\ &= 1-P(x<X, x<Y) \\ & = 1-P(X>x)\,P(Y>x)\\ & = 1-(1-P(X \leq x))\,(1-P(Y \leq x))\\ & = 1-(1-F_X(x))\,(1-F_Y(x))\end{split}$$
But now how

calculate expected value of random variable $Z=\min\left\{X,Y\right\}$

So we have $Z = 1-(1-F_X(x))\,(1-F_Y(x))$
Then $$E(Z) = \int_{0}^{\infty}\left(1-(1-F_X(x))\,(1-F_Y(x))\right) \,\,dx$$
Is it good like this?
 A: When $Z=\min(X,Y)$ and $X$ and $Y$ are independent is true to say that $F_Z(x)=1-(1-F_X(x))\,(1-F_Y(x))$  
Since $Z$ is a non-negative random variable here, you can say $\mathbb E[Z]=\int_0^\infty (1-F_Z(x)) \, dx$ which would be $$\int_0^\infty (1-F_X(x))\,(1-F_Y(x)) \, dx$$ here, but this is not quite what you have said.  Your expression looks to me more like $\int_0^\infty F_Z(x) \, dx$, which would be infinite.
Even easier here would be to note that $X,Y,Z$ are non-negative integers, so the integral is not going to be simple.  Instead, you can turn it into a sum, with $$\mathbb E[Z]=\sum_0^\infty (1-F_Z(n)) =\sum_0^\infty (1-F_X(n))\,(1-F_Y(n))$$
Here $F_X(n)=F_Y(n) = 1 - \dfrac1{2^{n}}$ so $$\mathbb E[Z]=\sum_0^\infty \frac1{4^{n}} =\dfrac43$$  
A: First, it is not correct to write  $Z = 1 - (1-F_X (x))(1 -F_Y (x))$. I think you mean that the cumulative distribution function of $Z$ is $F_Z(k) = 1 - (1-F_X(k))(1-F_Y(k))$. Then the probability density funcion $f_Z$ from $Z$ is given by 
$$f_Z(k) = F_Z(k) - F_Z(k-1)$$
Furthermore note that by your definition $Z$ is a discrete valued random variable, so it is not right to write it as a random variable with density function respectivly to $dx$.
The right calculation should be the sum:
$$ \Bbb{E}[Z] =\sum_{k=1}^\infty k f_Z(k)$$
