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?
Best Answer
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 $$\begin{align} \mathbb{P}\{ |Z| \leq 2a - a \} &\leq \mathbb{P}\{ |Z| \leq |\mathbb{E}[Z]| - a \} = \mathbb{P}\{ |\mathbb{E}[Z]| - |Z| \geq a \}\\ &\leq \mathbb{P}\{ |Z-\mathbb{E}[Z]| \geq a \}\\ &\leq \frac{\mathrm{Var}[Z]}{a^2} \tag{3} \end{align}$$