Timeline for Analog of Chebyshev's inequality for higher moments
Current License: CC BY-SA 2.5
2 events
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| Jun 16, 2010 at 0:25 | comment | added | Kevin O'Bryant | I remember Markov/Chebyshev like this: If $X$ is nonnegative, then the area above the distribution function $F(x)=\mathbb P(X \leq x)$, below 1, and to the right of $x=0$ is $\mathbb E[X]$. Since that region includes a rectangle with width $\lambda$ and height $1-\mathbb P(X\leq \lambda) = \mathbb P(X>\lambda)$, I get Markov's inequality: $E[X] \geq \lambda \mathbb P(X>\lambda)$. Apply to the correct nonnegative $X$ with the correct $\lambda$ to proceed to Chebyshev! | |
| Jun 15, 2010 at 23:07 | history | answered | Tom LaGatta | CC BY-SA 2.5 |