I have a concentration inequality like
For any
$\xi \geq 2$,
\begin{align}
\mathbb{P}\left(\left(|h_{w}\left(s,p\right)-\mathbb{E} h_{w}\left(s,p\right)|\right) \geq C_{\beta, D_{L S_\beta}} m(\xi,h_{w} )\right) \leq e^{-\xi}\\
\mathbb{P}\left(\underbrace{(|h_{w}\left(s,p\right)-\mathbb{E} h_{w}\left(s,p\right)|}_{Y} \geq C_{\beta, D_{L S_\beta}} m(\xi,h_{w} )\right) \leq e^{-\xi}\\
\mathbb{P}\left(Y\cdot z \geq C_{\beta, D_{L S_\beta}} m(\xi,h_{w} )\right) \leq e^{-\xi} &&\text{True}?
\end{align}
Reference
Now here's my question: If I multiply another random variable (say, $z$) with $Y$, where $z \leq 2B$, what happens? Will the above inequality holds? C is a constant
The main question is something like this** : If I multiply a random variable with another bounded random variable , will the concentration inequality change for the product or will it remain the same?
-## UPDATE ##
My $Y,z$ are non-negative random variables and $f(\xi) = e^{-\xi}$
$$\mathbb P(Y \geq C_{\beta, D_{L S_\beta}} m(\xi,h_{w} ) )\le f(\xi),\ \ \text{ and }\ \mathbb P(z \le 2B) =1 $$
We can write,
$$\{Yz \geq C_{\beta, D_{L S_\beta}} m(\xi,h_{w} ) \}\subseteq\{Y \cdot 2B \geq C_{\beta, D_{L S_\beta}} m(\xi,h_{w} )\} $$
$$\mathbb P(Yz \geq C_{\beta, D_{L S_\beta}} m(\xi,h_{w} ) )\le\mathbb P(Y\cdot 2B \geq C_{\beta, D_{L S_\beta}} m(\xi,h_{w} ))=\mathbb P\left(Y \geq \frac{C_{\beta, D_{L S_\beta}} m(\xi,h_{w} )} {2B}\right)\le f\left(\frac{\xi}{2B}\right) $$
\begin{align}\boxed{
\mathbb{P}\left(\left(|h_{w}\left(s,p\right)-\mathbb{E} h_{w}\left(s,p\right)| z \right) \geq C_{\beta, D_{L S_\beta}} m(\xi,h_{w} )\right) \leq e^{\frac{-\xi}{2B}}}
\end{align}
Is this proof okay? Or my boxed inequality seems wrong?
Best Answer
In general, if $X,Z$ are non-negative random variables and you know that for all $t>0$
$$\mathbb P(X>t)\le f(t),\ \ \text{ and }\ \mathbb P(Z\le M) =1\tag A$$ For some (deterministic) function $f$ and constant $0<M<\infty$, then using the identity $\Pr(A) = \Pr(A\cap B) + \Pr(A\cap B^c)$ for any two events $A$ and $B$, it holds that : $$\mathbb P(X>t) = \mathbb P(X>t,Z\le M) + \mathbb P(X>t, Z> M)$$ But notice that $ \mathbb P(X>t, Z> M) \le \mathbb P(Z> M)= 0$ hence $$ \mathbb P(X>t) = \mathbb P(X>t,Z\le M) $$ And by the exact same argument $$ \mathbb P(XZ>t) = \mathbb P(XZ>t,Z\le M) \tag B $$ Now observe the implication
$$\begin{align} \omega \in \left\{XZ>t\right\} \cap \left\{Z\le M\right\} &\implies X(\omega)Z(\omega)>t \text{ and } Z(\omega)\le M \\ &\implies X(\omega) M \ge X(\omega)Z(\omega)>t\\ &\implies \omega \in \{XM>t\} \end{align}$$
In other words you have the inclusion of events
$$\left\{XZ>t\right\} \cap \left\{Z\le M\right\}\subseteq\left\{XM>t\right\} \tag C$$
Which implies that for all $t>0$, $$\mathbb P(XZ>t) \stackrel{(B)}{=} \mathbb P(XZ>t,Z\le M)\stackrel{(C)}{\le}\mathbb P(XM>t)=\mathbb P\left(X>\frac t M\right)\stackrel{(A)}{\le} f\left(\frac t M\right) $$