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Here is a statement of Markov's inequality.

Suppose that $X$ is a random variable and that $g: \mathbb{R}\to [0,\infty]$ is Borel measurable and non-decreasing. Then, for any real $c$, $$ E[g(X)]\geq E[g(X):X \geq c] \geq g(c)P(X\geq c) \tag{1}$$

I like to understand how the assumptions on $g$ are used in (1).

Am I correct that

(a) the first inequality in (1) is true regardless of what $g$ is(that is, we don't need any assumption on g)?

(b) the second inequality in (1) is true because $g$ is non-decreasing?

But, then I do not understand why $g$ is assumed to be non-negative valued. Why do we need this? am I missing something?

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    $\begingroup$ Are you sure the first inequality would be true, if $\textrm{Range}(g) = [-\infty, 0]$? $\endgroup$ – TenaliRaman Oct 18 '15 at 19:36
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    $\begingroup$ (a) Nonnegativity of $g$: $g(X)\ge g(X)1\{X\ge c\}$ $\endgroup$ – d.k.o. Oct 18 '15 at 19:37
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    $\begingroup$ @TenaliRaman I now got it. I was mistaken to think that integrating over the "smaller" set $X\geq c$ is going to lead to a smaller value of integral for any $g$ $\endgroup$ – user74261 Oct 18 '15 at 19:44
  • $\begingroup$ @d.k.o. I now understand. I mistakenly thought that integrating over a smaller set gives a smaller value in any case $\endgroup$ – user74261 Oct 18 '15 at 19:45

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