Probability: Difference between revisions
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: <math>(g(X_n) - g(\theta)) \approx g'(\theta)(X_n - \theta)</math> | : <math>(g(X_n) - g(\theta)) \approx g'(\theta)(X_n - \theta)</math> | ||
==Limit Theorems== | ==Inequalities and Limit Theorems== | ||
===Markov's Inequality=== | ===Markov's Inequality=== | ||
Let <math>X</math> be a non-negative random variable.<br> | |||
Then <math>P(X \geq a) \leq \frac{E(X)}{a}</math> | |||
{{hidden | Proof | | |||
<math> | |||
E(X) | |||
= \int_{0}^{\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}af(x)dx | |||
=a \int_{a}^{\infty}f(x)dx | |||
=a*P(X \geq a)\\ | |||
\implies P(x\geq a) \leq \frac{E(X)}{a} | |||
</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> |