Calculating $E(\max\{X,Y\})$ using the tail theorem - is there a mistake in my solution? Given $X\sim \mathrm{Geom}(\frac 1 2), Y\sim \mathrm{Geom}(\frac 1 6)$ (starting with $1$, not $0$), I need to calculate $E(\max\{X,Y\})$ using the tail theorem.
The following picture shows my solution:

What bothers me, is that I wouldn't expect the answer to be less than $6$, which is the mean of $Y$.
What do you think? Is there a mistake in my solution or does this make sense somehow?
 A: Note that generally $$\max{\{X,Y\}}=X+Y-\min{\{X,Y\}}$$ This is useful here, because it is known (see here) that the minimum of two geometric random variables $X,Y$ with $p_X=1/2$ and $p_Y=1/6$ is again geometrically distributed with parameter $$p=1-(1-1/2)(1-1/6)=\frac{7}{12}$$
So, by linearity of the expectation $$E[\max\{X,Y\}]=E[X]+E[Y]-E[\min\{X,Y\}]=2+6-\frac{12}{7}=6.2857$$

Of course the same conclusion can be drawn with the tail theorem as you need it. Since the random variables are discrete you do not need the integral, but a sum instead. So, if $M=\max{\{X,Y\}}$ then 
\begin{align}\mathbb E[M]&=\sum_{k=0}^{+\infty}P(M> k)=\sum_{k=0}^{+\infty}(1-\mathrm P(M\leqslant k))=\sum_{k=0}^{+\infty}(1-\mathrm P(X\le k)P(Y\le k))\\[0.2cm]&=\sum_{k=0}^{\infty}(1-(1-P(X>k))(1-P(Y>k)))\\[0.2cm]&=\sum_{k=0}^{\infty}\left(1-\left(1-\frac1{2^k}\right)\left(1-\left(\frac5{6}\right)^k\right)\right)=\sum_{k=0}^{+\infty}\frac1{2^k}+\sum_{k=0}^{+\infty}\left(\frac5{6}\right)^k-\sum_{k=0}^{+\infty}\left(\frac5{12}\right)^k \\[0.2cm]&=\frac{1}{1-\frac12}+\frac{1}{1-\frac56}-\frac{1}{1-\frac5{12}}=2+6-\frac{12}7=6.2857\end{align}
A: For one thing, your formula for $F(x)$ is valid only when $x$ is an integer.  Because these are integer-valued random variables, $F$ is piecewise constant.  Consequently I'd re-write the integral as a discrete sum.
$$
F(x) = 1 - \left( \frac 5 6 \right)^{\lfloor x \rfloor} - \left( \frac 1 2 \right)^{\lfloor x\rfloor} + \left( \frac 5 {12} \right)^{\lfloor x\rfloor} \text{ for }x\ge 0.
$$
Since $\Pr(\max\{X,Y\}\ge Y)=1$ and $\Pr(\max\{X,Y\}> Y)>0$, certainly you have $\operatorname{E}(\max\{X,Y\})>\operatorname{E}(Y) = 6$.
The interval of constancy of $F(x)$ for $x\ge0$ is $[n,n+1)$ for $n\in\{0,1,2,\ldots\}$.  Then integral of a constant from $n$ to $n+1$ is just that constant.  So you have
$$ \int_0^\infty 1-F(x)\,dx = \sum_{n=0}^\infty (1-F(n)). $$
