----- > [!proposition] Proposition. ([[Chebyshev's Inequality]]) > Let $(X, \Sigma, \mu)$ be a [[measure|measure space]] [[Lp-norm|and]] $f \in \mathcal{L}^{p}(\mu)$ for $0 < p < \infty$. Then, for $t>0$, $\mu \left\{ |f| \geq t \right\} \leq \frac{\|f\|^{p}_{p}}{t^{p}} \, .$ ^119db4 > [!proof]+ > Apply [[Markov's inequality]] with $h=|f|^{p}$ and $c=t^{p}$ to obtain $\mu \{ |f|^{p} \geq t^{p} \} \leq \underbrace{ \| \ |f|^{p} \ \|_{1} }_{ \|f\|_{p}^{p} } / t^{p}.$ Upon recognizing $\mu \{ |f|^{p} \geq t^{p} \}=\mu \{ |f| \geq t \}$, we are done. ^proof-1 > [!specialization] Specialization. ([[Chebyshev's Inequality|Chebyshev's expectation inequality]]) > Suppose $(\Omega, \mathcal{F}, \mathbb{P})$ is a [[probability|probability space]] and $X \in \mathcal{L}^{2}(\mathbb{P})$ is a [[random variable]]. [[variance|If]] $\sigma(X)=0$ then we already know $X$ is constant a.s. So assume $\sigma(X)>0$. Then, for $t>0$, $\mathbb{P}\{ |X- \mathbb{E}X| \geq t\} \leq \frac{\operatorname{var }X}{t^{2}}.$ > (Put $p=2$ and $f:=X - \mathbb{E}X$ in the original proposition statement.) > Equivalently, $\mathbb{P} \{ |X-\mathbb{E}X| \geq t \sigma(X) \} \leq \frac{1}{t^{2}}.$ > [!proof] Proof of Equivalence. > To see that the first form implies the second, convert the first to the second by substituting $t \mapsto t \sigma(X)$ into the first form to obtain $\mathbb{P}\{ |X-\mathbb{E}X| \geq t \sigma(X) \} \leq \underbrace{ \frac{\operatorname{var }X}{t^{2} \sigma^{2}(X)} }_{ \frac{1}{t^2} }.$ To see that the second implies the first, convert the second into the first by substituting $t \mapsto \frac{t}{\sigma(X)}$ into the second form to obtain $\mathbb{P} \{ |X-\mathbb{E}X| \geq t \} \leq \frac{1}{\left( \frac{t^{2}}{\sigma^{2}(X)} \right)}. $ ^proof > [!proof]- Proof. ([[Chebyshev's Inequality]]) > Follows from [[Markov's inequality]]. ![[CleanShot 2022-11-11 at 20.48.20.jpg]] ----- #### #analysis/probability-statistics # Statement Let $X: \Omega \to \mathbb{R}$ be a [[random variable]] whose [[expectation]] and [[variance]] both exist. Then, for $a>0$, $P(\{\omega : |X(\omega) - E(X)| \geq a\}) \leq \frac{\text{var}(X)}{a^2}.$ # Proof Follows from [[Markov's inequality]]. ![[CleanShot 2022-11-11 at 20.48.20.jpg]] # Example ![[CleanShot 2022-11-11 at 20.48.43.jpg]] #notFormatted ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```