I've 2 random variables X,Y.
Assume $ X<= Y$.
Prove that for each $\epsilon$ > 0 there exists $\delta$ > 0 so that if:
$|E[Y] - E[X]| < \delta $
Then:
$P(Y>= X + \epsilon) < \epsilon$
Could use some help.
A question in expectation and markov inequality
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$\begingroup$
probability
expectation
2 Answers
3
By Markov's Inequality,$$P(Y\geq X+\epsilon)=P(|Y-X|\geq\epsilon)\leq \frac{E[Y-X]}{\epsilon}<\frac{\delta}{\epsilon}.$$ Choose $\delta<\epsilon^2$.
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Choose $\delta < \epsilon^2$,
\begin{align} P(Y \geq X+ \epsilon) &=P(Y-X \geq \epsilon) \\ &\leq \frac{E[Y-X]}{\epsilon} \\ &< \frac{\epsilon^2}{\epsilon} \\&=\epsilon \end{align}
In particular, you can choose $\delta = \frac{\epsilon^2}{2}$