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\section{Markov and Chebyshev}

\begin{theorem}
{\bf Markov:} Let $X$ be a random variable assuming only non-negative
values.  Then for all positive real numbers:
\[
P(X\geq t) \leq E(X)/t
\]
Equivalently
\[
P(X \geq kE(X)) \leq 1/k
\]
\end{theorem}

\emph{Proof} by contradiction.

For a random variable $X$ with expectation $\mu_{x}$, its variance
$\sigma^{2}_{x}$ is defined to be $E(X-\mu_{x}^{2})$


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