# Bounded in Probability proof

#### JohnK

##### New Member
Hi all,

I could use some help understanding the following proof so please feel free to pitch in!

Consider any random variable $$X$$ with distribution function $$F_X(x)$$. Then given $$\epsilon >0$$ we can bound $$X$$ in the following way. Because the lower limit of $$F_X(x)$$ is 0 and its upper limit is 1, we can find $$y_1\ and\ y_2$$ such that:

$$F_X(x)<\epsilon/2$$ for $$x\leq y_1$$ and $$F_X(x)>1-\epsilon/2$$ for $$x\geq y_2$$

Let $$y=max\left\{|y_1|,|y_2|\right\}$$ then

$$\displaystyle P[|X|\leq y]=F_X(y)-F_X(-y-0)\geq 1-\epsilon/2-\epsilon/2=1-\epsilon$$

What I need help understanding is where the inequality in the last line comes from precisely. I know it has something to do with the bounds we set up above but I cannot figure it. I am a beginner so please explain in detail. Thanks.

#### Mean Joe

##### TS Contributor
Note that
1a) $$y \geq y_1$$, so $$-y \leq -y_1$$
Thus,
2a) $$F_X(-y) < \epsilon/2$$, or equivalently $$-F_X(-y) > -\epsilon/2$$

On the other hand,
1b) $$y \geq y_2$$
Thus,
2b) $$F_X(y) > 1-\epsilon/2$$

#### JohnK

##### New Member
Note that
1a) $$y \geq y_1$$, so $$-y \leq -y_1$$
Thus,
2a) $$F_X(-y) < \epsilon/2$$, or equivalently $$-F_X(-y) > -\epsilon/2$$

On the other hand,
1b) $$y \geq y_2$$
Thus,
2b) $$F_X(y) > 1-\epsilon/2$$

Wow, thanks a lot. One thing though, in 1a) don't you mean $$-y\leq y_1$$? That is also implied by the inequality and I think this is the result we need.

Last edited:

#### BGM

##### TS Contributor
I think it should be like

$$-y \leq -|y_1| \leq y_1$$

#### BGM

##### TS Contributor
It is because

$$y = \max\{|y_1|, |y_2|\} \geq |y_1| \Rightarrow -y \leq -|y_1|$$