Understanding Chebychev's Inequality: Proof, Conditions, and Implications

[Hjensen]

I am having a bit of a problem with Chebychev's inequality which is:

[tex]P(|X-\mu |\geq \alpha )\leq \frac{\mathrm{Var}(X)}{\alpha ^2}[/tex]

For a positive [tex]\alpha[/tex]. Here X denotes a stochastic variable with mean [tex]\mu[/tex] and finite variance. I am asked to give a direct proof of this result, using the inequality

[tex]Z^2\geq \textbf{1}_{|Z|\geq 1}, \quad Z=\frac{X-\mu}{\alpha}[/tex].

I have solved this part of the problem already. Next, however, I am asked to give sufficient and necessary conditions for equality in Chebychev, and this is where I could use some help. My initial idea was to use the other inequality, but I am not quite sure how I can translate this into the Chebychev inequality. I figure that equality will be achieved for -1, 0, and 1 respectively, due to the nature of this characteristic function, but I should probably have further conditions on that. Any help would be appreciated.

 

[lippyka]

Thank you. The sufficient and necessary condition for equality in Chebychev's inequality is that the random variable X must be concentrated on a single value, i.e., it must have a Dirac delta distribution. In other words, X must take a single value with probability 1. This means that all moments of X must be 0, except the zeroth moment, which is 1. In this case, the inequality can be written as P(|X-\mu |\geq \alpha ) = \frac{\mathrm{Var}(X)}{\alpha ^2} = 1. This implies that the variance of X must be equal to \alpha^2. This is the necessary and sufficient condition for equality in Chebychev's inequality.