# Tuesday February 18th

**Markov's inequality:**

\begin{align*}
P(X\geq a) \leq \frac{ E(X) }{a} \\
P(X\geq a) \leq \frac{ E(\abs X) }{a} \\
P(X^2 \geq a) \leq \frac{ E(X^2) }{a} \\
.\end{align*}

For any sequence $X_n\geq 0$, the sequence $S_n = \sum_{i=1}^n X_i$ is monotone increasing, so

\begin{align*}
E(\sum X_n) = \sum E(X)
.\end{align*}

**Proposition:**
For $b(x)$ a strictly increasing positive function, for any random variable $X\geq 0$ we have

\begin{align*}
\sum_{n=1}^\infty P(X \geq b(n)) \leq E b\inv(X) \leq \sum_{n=0}^\infty P(X > b(n))
.\end{align*}

Note that $\lim_n \int f_n \neq \int \lim_n f_n$ in general.
The following condition is necessary and sufficient for convergence:

**Definition:**
A sequence of random variables $X_n$ is *uniformly integrable* iff 

1. $\forall\eps>0$ there exists a $\delta > 0$ such that $\forall A\in F$, $P(A) < \delta \implies \sup_n \int_A \abs{X_n} ~dp < \eps$
2. $\sup_n E\abs{X_n} < \infty$.

**Proposition:**
If $X_n^{\pm}$ (positive/negative parts respectively) are uniformly integrable, then $\theset{X_n}$ is uniformly integrable from above/below.

**Lemma (Criterion for Integrability):**
A measurable function $X$ is integrable $\iff$ for all $\eps,~\exists \delta$ such that

\begin{align*}
P(A) < \delta \implies \int_A \abs{X} dp < \eps, \quad E\abs{X} \leq \frac 1 \delta
.\end{align*}

**Proposition (Criterion for Uniform Integrability):**
A sequence of random variables $\theset{X_n}$ is uniformly integrable iff 
$$
\lim_{a\to\infty}\sup_n \int_{\abs{X_n} > a} \abs {X_n} ~dp = 0
.$$