# Almost sure convergence via pairwise independence

If ${A_1,A_2,...}$ are pairwise independent and ${\sum_{n=1}^{\infty}P(A_n)=\infty}$ then as ${n \rightarrow \infty}$

$\displaystyle \boxed{ \frac{\sum_{m=1}^{n}\mathbb{I}_{A_m}}{\sum_{m=1}^{n}P(A_m)} \xrightarrow{a.s.} 1 }$

Proof:

Let ${X_m = \mathbb{I}_{A_m}}$ and ${S_n = X_1+...+Xn}$. Since ${A_m}$ are pairwise independent, the ${X_m}$ are uncorrelated and thus

$\displaystyle var(S_n) = var(X_1) + ... + var(X_n)$

Since ${X_m \in \{0,1 \}}$

$\displaystyle var(X_m) \leq \mathbb{E}[X_m^2] = \mathbb{E}[X_m] \Rightarrow var(S_n) \leq \mathbb{E} [S_n]$

# An inequality of the mean involving truncation

Let ${X_1,X_2,...}$ be i.i.d. r.vs with ${\mathbb{E}[|X_i|] < \infty}$ and ${Y_k = X_k \mathbb{I}_{(|X_k| \leq k)}}$. Then

$\displaystyle \boxed{ \mathbb{E}[X_1] \geq \sum_{k=1}^{\infty} \frac{var(Y_k)}{4 k^2 } }$

Proof:

First we proove the following useful result

If ${X \geq 0}$ and ${ a > 0}$ then

$\displaystyle \boxed{ \mathbb{E}[X^a] = \int_{0}^{\infty} a x^{a-1} P(X >x) dx }$

$\displaystyle \begin{array}{rcl} \int_{0}^{\infty} a x^{a-1} P(X >x) dx &=& \int_{0}^{\infty} \int_{\Omega}a x^{a-1} \mathbb{I}_{(X>x)} dP dx \\ &=& \int_{\Omega} \int_{0}^{\infty} a x^{a-1} \mathbb{I}_{(X>x)} dP dx \\ &=& \int_{\Omega} \int_{0}^{X} a x^{a-1} dP dx = \mathbb{E}[X^a] \end{array}$

Note you can find the same lemma on Feller Vol.2 (p. 150) as

$\displaystyle \mathbb{E}[X^a]= \int_{0}^{\infty} x^a F \{ dx \} = a \int_{0}^{\infty} x^{a-1} [ 1- F(x)] dx$

# Some notes on Kalman Filtering

State Space form

Measurement Equation

$\displaystyle \boxed{\mathbf{\underbrace{y_{t}}_{N \times 1}=\underbrace{Z_{t}}_{N \times m}\underbrace{a_{t}}_{m \times 1}+d_{t}+\varepsilon_{t}}}$

$\displaystyle Var(\varepsilon_{t})= \mathbf{H_{t}}$

Transition Equation

$\displaystyle \boxed{\mathbf{\underbrace{a_{t}}_{m \times 1} =\underbrace{T_{t}}_{m \times m} a_{t-1}+c_{t}+\underbrace{R_{t}}_{m \times g} \underbrace{\eta_{t}}_{g \times 1}}}$

$\displaystyle Var(\eta_{t})=\mathbf{Q}_{t}$

$\displaystyle E(a_{0})= \mathbf{a_{0} \; \; \; \; Var(a_{0})=P_{0}} \; \; \; \; E(\varepsilon_{t}a_{0}^{\top}) \; \; \; E(\eta_{t}a_{0}^{\top})$

Future form

$\displaystyle \mathbf{a_{t+1}=T_{t}a_{t}+c_{t}+R_{t}\eta_{t}}$

# Big Data for Volatility vs.Trend

So different aspects of Big Data — in this case dense vs. tall — are of different value for different things.  Dense data promote accurate volatility estimation, and tall data promote accurate trend estimation.

More (No Hesitations blog)

# Rio’s Inequality

Let ${X}$ and ${Y}$ be two integrable real-valued random variables and let ${ Q_x(u) = inf\{t: P(|X|>t) \leq u \}}$ be the quantile function of ${|X|}$. Then if ${Q_X Q_Y}$ is integrable over ${ (0,1)}$ we have

$\displaystyle \boxed{|Cov(X,Y)| \leq 2 \int\limits_{0}^{2a} Q_x(u) Q_Y(u) du}$

where ${ a= a(\sigma(X), \sigma(Y)) = \sup\limits_{\substack{B \in \mathcal{B} \\ C \in \mathcal{C}}} |Cov(\mathbb{I}_{\sigma(X)},\mathbb{I}_{\sigma(Y)})|}$ is the a-mixing coefficient.

Proof: Set ${X^{+} = sup(0,X)}$ and ${X^{-} = sup(0,-X)}$ then

$\displaystyle Cov(X,Y) = Cov(X^{+},Y^{+}) + Cov(X^{-},Y^{-}) - Cov(X^{+},Y^{-}) - Cov(X^{-},Y^{+})$

since ${ X = (X^{+} - X^{-})}$ and ${ Y = (Y^{+} - Y^{-})}$

note also that

$\displaystyle Cov(X^+,Y^+) = \int \int_{\mathbb{R}^{2}_{+}} [P(X>u, Y> \upsilon) - P(X>u)P(Y> \upsilon)]du d\upsilon$

which implies that

$\displaystyle |Cov(X^+,Y^+)| \leq \int \int_{\mathbb{R}^{2}_{+}} \inf (a, P(X>u),P(Y> \upsilon))du d\upsilon$

# An inequality of the expectation

Let ${X_1,X_2,...}$ be i.i.d. r.vs with ${\mathbb{E}[|X_i|] < \infty}$ and ${Y_k = X_k \mathbb{I}_{(|X_k| \leq k)}}$. Then

$\displaystyle \boxed{ \mathbb{E}[X_1] \geq \sum_{k=1}^{\infty} \frac{var(Y_k)}{4 k^2 } }$

Proof:

$\displaystyle \begin{array}{rcl} var(Y_k) \leq \mathbb{E}[Y_k^2] & =& \int_{0}^{\infty}2y P(|Y_k|>y)dy \\ &\leq& \int_{0}^{k}2 y P(|X_1|>y)dy \end{array}$

So

$\displaystyle \begin{array}{rcl} \sum_{k=1}^{\infty} \mathbb{E}[Y_k^2]/k^2 &\leq& \sum_{k=1}^{\infty} k^{-2} \mathbb{I}_{(yy ) dy \\ &=& \int_{0}^{\infty} \left\lbrace \sum_{k=1}^{\infty} k^{-2} \mathbb{I}_{(yy ) dy \end{array}$