Probability: Difference between revisions
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{{hidden | Proof | | {{hidden | Proof | | ||
<math> | <math> | ||
\begin{aligned} | |||
E(X) | E(X) | ||
= \int_{0}^{\infty}xf(x)dx | &= \int_{0}^{\infty}xf(x)dx \\ | ||
= \int_{0}^{a}xf(x)dx + \int_{a}^{\infty}xf(x)dx | &= \int_{0}^{a}xf(x)dx + \int_{a}^{\infty}xf(x)dx\\ | ||
\geq \int_{a}^{\infty}xf(x)dx | &\geq \int_{a}^{\infty}xf(x)dx\\ | ||
\geq \int_{a}^{\infty}af(x)dx | &\geq \int_{a}^{\infty}af(x)dx\\ | ||
=a \int_{a}^{\infty}f(x)dx | &=a \int_{a}^{\infty}f(x)dx\\ | ||
=a*P(X \geq a)\\ | &=a * P(X \geq a)\\ | ||
\implies P( | \implies& P(X \geq a) \leq \frac{E(X)}{a} | ||
\end{aligned} | |||
</math> | </math> | ||
}} | }} | ||
===Chebyshev's Inequality=== | ===Chebyshev's Inequality=== | ||
* <math>P(|X - \mu| \geq k \sigma) \leq \frac{1}{k^2}</math> | * <math>P(|X - \mu| \geq k \sigma) \leq \frac{1}{k^2}</math> | ||