Given a scalar random variable $X$, suppose that there are positive constants $c_{1}$ and $c_{2}$ such that $$\forall t\geq 0 : \,\,\,\,\,\,\ \mathbb P\{|X-\mathbb EX|\geq t\}\leq c_{1}e^{-c_{2}t^{2}}$$ A median $m_{X}$ is any number such that $\mathbb P \{ X \geq m_{X} \} \geq \frac{1}{2}$ and $\mathbb P \{ X \leq m_{X} \} \geq \frac{1}{2}$. Show that for any median $m_{X}$, we have $$\mathbb P \{ |X-m_{X}| \geq t \} \leq 4c_{1}e^{-\frac{1}{8}c_{2}t^{2}}.$$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
5
Letting $t=0$ in your first displayed inequality, note that necessarily $c_1\ge1$. Let then $u<EX$ be the root of the equation $$c_1e^{-c_2(EX-u)^2}=1/2. $$ Then $$P(X<u)=P(X-EX<u-EX)\le P(|X-EX|>EX-u)\le c_1e^{-c_2(EX-u)^2}=1/2, $$ so that $m:=m_X\ge u$. Take any real $t\ge0$. If $$EX-u\le t/2,$$ then $$P(X-m\ge t)=P(X-EX\ge m-EX+t)\le P(X-EX\ge u-EX+t)\le P(X-EX\ge t/2)\le c_1e^{-c_2t^2/4}\le2c_1e^{-c_2t^2/4}. $$ On the other hand, if $$EX-u>t/2,$$ then $$P(X-m\ge t)\le1=2c_1e^{-c_2(EX-u)^2}\le2c_1e^{-c_2t^2/4}. $$ So, $$P(X-m\ge t)\le2c_1e^{-c_2t^2/4} $$ in any case. Similarly, $P(-(X-m)\ge t)\le2c_1e^{-c_2t^2/4}$. Thus, $$P(|X-m|\ge t)\le4c_1e^{-c_2t^2/4}. $$ This is a bit better than what you wanted (with $t^2/8$ in place of $t^2/4$).
answered Apr 3, 2019 at 21:39
-
$\begingroup$ Thank you. The problem has a similar converse: Show that whenever for a median $m_{X}$ we know $$\mathbb P \{ |X-m_{X}| \geq t \} \leq c_{3}e^{-c_{4}t^{2}}$$ then $$\mathbb P\{|X-\mathbb EX|\geq t\}\leq 2c_{3}e^{-\frac{1}{4}c_{4}t^{2}}.$$ $\endgroup$– MeysamCommented Apr 5, 2019 at 16:55
-
$\begingroup$ Actually the existence of such a $u$ should be proved and it is clear that, it does not conclude from $c_{1} \geq1$ and we should prove $\ln (2c_{1}) >0$. It can be proved like this: The problem in the case that $\mathbb EX=m_{X}$ is trivial. So we assume that $\mathbb EX \neq m_{X}$. Choose a real number $\epsilon$ such that $\epsilon < |\mathbb E-m_{X}|$. Now we consider two cases: $\endgroup$– MeysamCommented Apr 11, 2019 at 5:37
-
$\begingroup$ 1. $\mathbb EX < m_{X}$: $$\frac{1}{2}\leq \mathbb P \{X \geq m_{X} \} \leq \mathbb P \{X-\mathbb EX \geq \epsilon \} \leq \mathbb P \{|X-\mathbb EX| \geq \epsilon \} \leq c_{1}e^{-c_{2}\epsilon^{2}}$$ So $$2c_{1} \geq e^{c_{2}\epsilon^{2}} \implies \ln 2c_{1} > 0$$ $\endgroup$– MeysamCommented Apr 11, 2019 at 5:38
-
$\begingroup$ 2. $\mathbb EX > m_{X}$: $$\frac{1}{2}\leq \mathbb P \{X \leq m_{X} \} \leq \mathbb P \{-X+\mathbb EX \geq \epsilon \} \leq \mathbb P \{|X-\mathbb EX| \geq \epsilon \} \leq c_{1}e^{-c_{2}\epsilon^{2}}$$ So $$2c_{1} \geq e^{c_{2}\epsilon^{2}} \implies \ln 2c_{1} > 0$$ $\endgroup$– MeysamCommented Apr 11, 2019 at 5:44
-
$\begingroup$ @Meysam : That the root $u$ exists (and is unique) is trivial. Indeed, this unique root $u$ equals $EX-\sqrt{\ln(2c_1)/c_2}$. Note also that $\ln(2c_1)\ge\ln2>0$. $\endgroup$ Commented Apr 11, 2019 at 11:41