Bounded in Probability proof

JohnK

New Member
#1
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
#2
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) \(F_X(y) > 1-\epsilon/2\)

Add (2a) and (2b)
 

JohnK

New Member
#3
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) \(F_X(y) > 1-\epsilon/2\)

Add (2a) and (2b)
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.
 
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