# Markov's inequality over sum of two functions of a RV

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.

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.
• 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. – pkenneth81 Mar 28 at 7:52
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 $$f_1(X)+f_2(X)$$.
• 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? – pkenneth81 Mar 28 at 7:56