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If I am given some random variable X that has a value of at least -6 and expectation of 0, I can trivially deduce the following. $E[X] = 0$ and $X\geq -5$. However, how would I go about finding the probability that $X \geq 10$ with Markov's inequality?

I have the following deduced so far. $$0 = E[X]= \int_{-5}^{10} XP(x)dx + \int_{10}^{\infty} XP(x)dx$$ where $P(x)$ represents a probability function, but I am clueless from here. I very well know that Markov's should only be applied to non-negative numbers, but the problem constraints dictate that it is possible for a value to be found.

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Hint: Do a change of random variable to get non-negativity. Set $Y=X+6$, so that $\mathbb{E}Y = 6$ and $Y \geq 0$ a.s.

Then, by Markov's inequality, $$\mathbb{P}\{X \geq 10\} = \mathbb{P}\{Y \geq 16\} \leq \frac{\mathbb{E}Y}{16} = \frac{3}{8}.$$

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