Given $P(X=0)=1-\frac1{a^2}$ and $P(X=a)=P(X=-a)\frac1{2a^2}$, compute $P(|X| \geq a)$

Question

Given is random variable $X:\begin{pmatrix}-a&0&a\\\frac{1}{2a^2}&1-\frac{1}{a^2}&\frac{1}{2a^2}\end{pmatrix}$. Calculate $P(|X| \geq a)$

I'm not sure how to do this correct? But that notation reminds me very much of Markov's inequality which says in general that $$P(X \geq a) \leq \frac{E(X)}{a}$$

So I would start by calculating the expected value of our given random variable $|X|$:

$$E(|X|) = \frac{1}{2a^2} \cdot |-a| + \left(1-\frac{1}{a^2}\right) \cdot |0| + \frac{1}{2a^2} \cdot |a| = \frac{|-a|}{2a^2}+0+\frac{|a|}{2a^2} = \frac{a}{2a^2}+\frac{a}{2a^2} = \frac{1}{a}$$

So we have that $$P(|X| \geq a) \leq \frac{\frac{1}{a}}{a} \Leftrightarrow P(|X| \geq a) \leq \frac{1}{a^2}$$

Is it really correct like that? I need to know it please because I would do it like that in the exam too? :s

Answer 1

From the fact that $\frac1{2a^2}$ shows up in your question we conclude that $a\neq0$. Now we discern in this context two cases:

• $a<0$ then $P(|X|\geq a)=1$
• $a>0$ then $P(|X|\geq a)=P(|X|\neq0)=\frac1{a^2}$

The first is based on the fact that $|X|$ is a non-negative random variable so that $P(|X|\geq a)=1$ for every $a<0$.

The second is based on the data in your question.

Answer 2

Markov's inequality only gives you an (unsharp) upper bound. The question asks for a value.

$$\begin{aligned} P(|X|\ge a) &= P(X = a) + P(X = -a) \\ &= \frac{1}{2a^2} + \frac{1}{2a^2} \\ &= \frac{1}{a^2} \end{aligned}$$

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Edit in response to drhab's comment.

If $a < 0$, the event $\{|X| \ge a\} = \{|X| \ge 0\}$ is always realised, so $P(|X| \ge a) = 1$.