Derivation of Markov's Inequality I have some problems understanding the derivation of Markov's inequality.
$$
\begin{align}
EX &= \int_{-\infty}^{\infty} xf_X(x)\text{d}x \\
&= \int_{0}^\infty xf_X(x)\text{d}x & \text{since } X\gt 0\\
&\geq \int_{a}^\infty xf_X(x)\text{d}x &\text{ for any }a \gt 0\\
&\geq\int_{a}^\infty af_X(x)\text{d}x & \text{since } x > a\text{ in the integrated region.}
\end{align}
$$
I don't understand the last step. Didn't we already use the fact that we were integrating over a larger region to write the inequality in the third line? Can someone help me with the difference? 
 A: We could be more verbose in the third step as follows: Let $a > 0$. Then
$$\int_0^\infty x f_X (x) \text{d} x = \int_a^\infty x f_X (x) \text{d} x + \int_0^a x f_X (x) \text{d} x.$$
Now, since $f_X(x) \geq 0$ and $x\geq 0 $ for $x\in [0,a]$ we have that $\int_0^a x f_X (x) \text{d} x \geq 0$ (This is where integrating over a larger/smaller region with a postitive integrand plays a role). Therefore
$$\int_a^\infty x f_X (x) \text{d} x + \int_0^a x f_X (x) \text{d} x \geq \int_a^\infty x f_X (x) \text{d} x + 0 = \int_a^\infty x f_X (x) \text{d} x$$
Now let us take a look on the last step: Here we want to use that for $x\in[a,\infty )$ we have $x \geq a$ and therefore it holds that $x f_X (x) \geq a f_X (x)$ (Here we just used that over our integral region we have a lower bound of the integrand). Altogether
$$\int_a^\infty x f_X (x) \text{d} x \geq \int_a^\infty a f_X (x) \text{d} x = a \int_a^\infty f_X (x) \text{d} x$$
A: $ f \geq g$ implies $\int f \geq \int g$. Integrate $x f(x) I_{(a,\infty)} \geq a f(x) I_{(a,\infty)}$ to get the last step.
A: The last step applies: $$\forall x\;[0\leq u(x)\leq v(x)]\implies\int u(x)\;dx\leq\int v(x)\;dx$$ This for $u(x)=af_X(x)\mathbf1_{(a,\infty)}(x)$ and $v(x)=xf_X(x)\mathbf1_{(a,\infty)}(x)$ where $a\geq0$.
A: The last step uses the fact that the function 
\begin{align} g : [a,\infty) \rightarrow [0, \infty)\\
 x \mapsto x f_X(x)
\end{align}
is always larger than
\begin{align} h : [a,\infty) \rightarrow [0, \infty)\\
 x \mapsto a f_X(x)
\end{align}
for (a > 0) in its domain. As you correctly state, the proof uses the fact that you integrate a non-negative function over a smaller region in the second-to-last inequality, not in the last inequality.
A: No, there is a difference. Second to third step uses the fact that
$\int_0^\infty xf_X(x)dx = \int_0^a xf_X(x)dx+\int_a^\infty xf_X(x)dx$
And
$\int_0^a xf_X(x) dx>0$
Last step uses that $\int_a^\infty xf_X(x)dx>\int_a^\infty af_X(x)dx$, because $x>a$ in these limits of integration 
Hope it is helpful
