Probability: Difference between revisions

No edit summary
Line 90: Line 90:
{{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(x\geq a) \leq \frac{E(X)}{a}
\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>