I'm trying to solve a problem that I think uses Markov Inequality(or chebyshev). But not sure how to apply it. the problem is like this:
Show that $$P(|X-E(X)|+|Y-E(Y)|\geq4\sigma)\leq\frac{1}{2}.$$ (the random variables X, Y have common variance $\sigma^2$)
I am assuming this is somehow extension of univariate case into multivariate, but I am not sure how to approach this question. I did prove the case where $P(|(X-E(X))+(Y-E(Y))|)\leq\frac{1+\rho}{2}$ But not sure about this case. Thank you.
Let $A=X-E(X)$ and $B=Y-E(Y)$. Then note that $$ |A| + |B| \ge 4\sigma \implies |A|>2\sigma \text{ or } |B|>2\sigma.$$ The same thing written with set notation: $$ \{ |A| + |B| \ge 4\sigma \} \subset \{ |A|>2\sigma\} \cup \{ |B|>2\sigma \}.$$ Applying monotonicity of $P$, union bound, and then Chebyshev's inequality gives $$ P(|A| + |B| \ge 4\sigma ) \le P(|A| \ge 2\sigma) + P(|B|\ge 2\sigma) \le\frac1{2^2} + \frac1{2^2} = \frac12. $$