mjhjmtu

I have some problems understanding the derivation of Markov's inequality.

$$
beginalign
EX &= int_-infty^infty xf_X(x)textdx \
&= int_0^infty xf_X(x)textdx & textsince Xgt 0\
&geq int_a^infty xf_X(x)textdx &text for any a gt 0\
&geqint_a^infty af_X(x)textdx & textsince x > atext in the integrated region.
endalign
$$

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?

We could be more verbose in the third step as follows: Let $a > 0$. Then
$$int_0^infty x f_X (x) textd x = int_a^infty x f_X (x) textd x + int_0^a x f_X (x) textd x.$$
Now, since $f_X(x) geq 0$ and $xgeq 0 $ for $xin [0,a]$ we have that $int_0^a x f_X (x) textd x geq 0$ (This is where integrating over a larger/smaller region with a postitive integrand plays a role). Therefore
$$int_a^infty x f_X (x) textd x + int_0^a x f_X (x) textd x geq int_a^infty x f_X (x) textd x + 0 = int_a^infty x f_X (x) textd x$$

Now let us take a look on the last step: Here we want to use that for $xin[a,infty )$ 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
$$int_a^infty x f_X (x) textd x geq int_a^infty a f_X (x) textd x = a int_a^infty f_X (x) textd x$$

$ 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.

The last step applies: $$forall x;[0leq u(x)leq v(x)]impliesint u(x);dxleqint v(x);dx$$ This for $u(x)=af_X(x)mathbf1_(a,infty)(x)$ and $v(x)=xf_X(x)mathbf1_(a,infty)(x)$ where $ageq0$.

The last step uses the fact that the function
beginalign g : [a,infty) rightarrow [0, infty)\
x mapsto x f_X(x)
endalign
is always larger than
beginalign h : [a,infty) rightarrow [0, infty)\
x mapsto a f_X(x)
endalign
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.

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