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?


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

  • $\begingroup$ 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
  • 1
    $\begingroup$ @Babado You're welcome! $\endgroup$
    – Clement C.
    Mar 11 '21 at 19:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.