- #1
Hjensen
- 23
- 0
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.
[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.