Derivation of Markov's Inequality

Question

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?

Answer 1

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 $$\int_a^\infty x f_X (x) \text{d} x \geq \int_a^\infty a f_X (x) \text{d} x = a \int_a^\infty f_X (x) \text{d} x$$

Answer 2

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.

Answer 3

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 $a\geq0$.

Answer 4

$ 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.

Answer 5

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