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I'm wondering whether Markov's inequality can be applied over the following example, as I need an upper bound for the probability determined by:

$ P( f_1(X) + f_2(X) \geq \alpha ) $

Above, X is a random variable with pmf $p_X$, and functions $f_1$ and $f_2$ are deterministic and well defined. (though, we don't know them)

May I just apply Markov's inequality and do:

$ P( f_1(X) + f_2(X) \geq \alpha ) \leq \frac{E_X[f_1(X) + f_2(X) ]}{\alpha} $

My intuition is telling me this is possible. But I may be missing something I don't know.

Thank you all for your comments.

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If you can guarantee $f_1(x)+f_2(x) \geq 0$ for all values $x$ that $X$ might attain, then yes. Define the random variable $Y=f_1(X)+f_2(X)$, which is non-negative, and apply Markov's inequality as usual: for $\alpha>0$, $P[Y>\alpha]\leq \mathbb{E}[Y]/\alpha$.

If you can't guarantee $f_1(x)+f_2(x) \geq 0$ then Markov's inequality won't hold.

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  • $\begingroup$ Thanks for your reply. I see, I think I was missing the fact that for Markov's inequality: $P(X>a) \leq \frac{E[X]}{a}$, one needs that X to be non-negative random variable to hold. I see your point and how well you addressed it. $\endgroup$ – pkenneth81 Mar 28 at 7:52
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The inequlaity is true if $f_1,f_2$ are measurable, $f_1(X)+f_2(X) \geq 0$ and $\alpha >0$. It is nothing but Markov's in equality applied to the random variable $f_1(X)+f_2(X)$.

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  • $\begingroup$ Thanks for your reply. So, lets say, a slightly different example: For $P(f_1(X)-f_2(x) \geq \alpha ) $ and both $f_1(X) >0$ and $f_2(X)>0$, Markov's inequality will not hold as we cannot guarantee $f_1(X)-f_2(x) $ is a non negative random variable. Am I right? $\endgroup$ – pkenneth81 Mar 28 at 7:56
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    $\begingroup$ @htennek2k You are right. Non-negativity is essential. $\endgroup$ – Kavi Rama Murthy Mar 28 at 8:06

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