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Let $Z = X + Y i$ be a complex random variable, where $X, Y$ are real-valued r.v.'s

Denote by $\mathbb{E}[Z]$ and $\sigma^2 = Var[Z]$ the expectation and variance of $Z$.

By Chebyshev's inequality,

$$P(|Z-\mathbb{E}[Z]| \geq \sigma k) \leq \frac{1}{k^2}$$

If $\mathbb{E}[|Z|] \geq 2 \sigma k$, is it true that

$$P(|Z| \leq \sigma k) \leq \frac{1}{k^2} ? $$

Why am I asking this?
If $Z$ was a real-valued r.v. and $E[Z] \geq 2 \sigma k $ then $P(Z \leq \sigma k) = P(Z \leq 2\sigma k - \sigma k) \leq P(Z \leq \mathbb{E}[Z] - \sigma k) \leq \underbrace{P(|Z - \mathbb{E}[Z]| \geq \sigma k) \leq \frac{1}{k^2}}_{Chebyshev}$.
For my question above, we cannot do exactly the same, as $| \ \cdot \ | $ doesn't have the same meaning in the complex numbers...
I think the answer to my question should be yes! but I'm missing something... Anyone can help?

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First, a detour to recap: the proof of Chebyshev's inequality is just using Markov's inequality on $X^2$, for
$X := |Z-\mathbb{E}[Z]|^2$.

$X$ is a non-negative, real-valued random variable, so $$ \mathbb{P}\{ |Z-\mathbb{E}[Z]| \geq a \} = \mathbb{P}\{ X \geq a^2 \} \leq \frac{\mathbb{E}[X]}{a^2} = \mathbb{P}\{ X \geq a^2 \} \leq \frac{\mathbb{E}[|Z-\mathbb{E}[Z]|^2]}{a^2} = \frac{\mathrm{Var}[Z]}{a^2} \tag{1} $$ still holds.

So we still have Chebyshev's inequality, as you had stated. (Just for the sake of it, it's worth restating.)

Now, for the second (main) part: the modulus does satisfy the triangle inequality, which is what we need: $$ |\mathbb{E}[Z]| - |Z| \leq ||\mathbb{E}[Z]| - |Z|| \leq |\mathbb{E}[Z]- Z| \tag{2} $$

If $|\mathbb{E}[Z]| \geq 2a$, then this implies $$\begin{align} \mathbb{P}\{ |Z| \leq 2a - a \} &\leq \mathbb{P}\{ |Z| \leq |\mathbb{E}[Z]| - a \} = \mathbb{P}\{ |\mathbb{E}[Z]| - |Z| \geq a \}\\ &\leq \mathbb{P}\{ |Z-\mathbb{E}[Z]| \geq a \}\\ &\leq \frac{\mathrm{Var}[Z]}{a^2} \tag{3} \end{align}$$

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  • $\begingroup$ I agree with what you wrote but I really did mean to assume $\mathbb{E}[|Z|] \geq 2a$ and not $|\mathbb{E}[Z]| \geq 2a$. Can we conclude something based on the first bound? $\endgroup$
    – Babado
    Mar 11 '21 at 0:54
  • $\begingroup$ @Babado Is that even true when $Z$ is real-valued? I think I can build a counterexample, e.g., $Z$ being $0$ w.p. $1/2$, $1$ w.p. $1/4$, and $-1$ w.p. 1/4. I can check after lunch, but then $\mathbb{E}[|Z|] = 1/2$, $\sigma = 1/\sqrt{2}$, and $\mathbb{P}\{ |Z| \leq \sigma k\} = 1/2$ for all $k\in(0, \sqrt{2})$. $\endgroup$
    – Clement C.
    Mar 11 '21 at 1:07
  • $\begingroup$ How is that a counterexample? If $k \leq \frac{\sqrt{2}}{4}$ then $\mathbb{E}[|Z|] \geq 2 \sigma k $. $\endgroup$
    – Babado
    Mar 11 '21 at 9:24
  • $\begingroup$ @Babado Take $k=\frac{1}{2\sqrt{2}}$. Then $2\sigma k = \frac{1}{2} = \mathbb{E}[|Z|]$, so your assumption is satisfied. But $$\frac{1}{2}=\mathbb{P}\{|Z| \leq \sigma k\}> \frac{1}{8} = \frac{1}{k^2}$$ so the conclusion doesn't. $\endgroup$
    – Clement C.
    Mar 11 '21 at 10:16
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    $\begingroup$ @Babado You're welcome! $\endgroup$
    – Clement C.
    Mar 11 '21 at 19:59

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