Welcome to the Ultimate Guide to Tennis W15 Campulung Romania
Join us as we dive into the thrilling world of the Tennis W15 Campulung Romania tournament. This event is not just a showcase of top-tier tennis talent but also a hub for expert betting predictions and daily match updates. Whether you're a seasoned tennis enthusiast or new to the game, this guide will keep you informed and entertained.
What is Tennis W15 Campulung Romania?
The Tennis W15 Campulung Romania is part of the Women's World Tennis Tour, offering a platform for emerging talents to shine on an international stage. Held in the picturesque town of Campulung Moldovenesc, this tournament brings together players from around the globe, competing for prestige and prize money. With matches updated daily, fans can follow every serve, volley, and point as the action unfolds.
Daily Match Updates
Stay ahead with our comprehensive daily match updates. We provide detailed coverage of each game, including player statistics, match highlights, and key moments. Our team of expert commentators ensures you don't miss a thing, offering insights and analysis that enhance your viewing experience.
- Live Scores: Get real-time updates on scores as they happen.
- Match Highlights: Watch replays of crucial points and rallies.
- Player Stats: Access in-depth statistics for each player.
Expert Betting Predictions
Betting enthusiasts will find our expert predictions invaluable. Our analysts use a combination of statistical analysis, player form, and historical data to provide accurate forecasts. Whether you're placing bets on the outcome of individual matches or the overall tournament winner, our insights can help you make informed decisions.
- Predictions: Daily forecasts for each match.
- Odds Analysis: Understanding the odds to maximize your bets.
- Tips from Experts: Advice from seasoned bettors.
Meet the Players
The Tennis W15 Campulung Romania features a diverse lineup of players, each bringing their unique style and skill set to the court. Get to know some of the standout athletes competing in this year's tournament:
- Jana Fett: Known for her powerful serve and aggressive playstyle.
- Alexandra Cadanțu: A formidable opponent with exceptional baseline skills.
- Nicoleta Dascălu: Renowned for her agility and quick reflexes.
Tournament Schedule
The tournament runs over several days, with matches scheduled throughout the day to accommodate fans across different time zones. Here's a snapshot of what to expect:
- Day 1: Opening matches featuring top-seeded players.
- Day 2-3: Quarterfinals and semifinals action intensifies.
- Day 4: The grand finale where champions are crowned.
Venue Highlights
The Campulung Moldovenesc sports complex provides an ideal setting for this prestigious event. With state-of-the-art facilities and a passionate local crowd, it offers an electrifying atmosphere for both players and spectators.
- Court Conditions: Fast-paced surfaces that challenge even the best players.
- Audience Experience: Comfortable seating and excellent views from every angle.
- Amenities: Top-notch facilities including food stalls, merchandise shops, and rest areas.
How to Watch
You won't miss a moment of the action with multiple viewing options available:
- Livestreams: Watch live matches online via our official website or partner platforms.
- Social Media Updates: Follow us on social media for real-time updates and exclusive content.
- Broadcast Partners: Tune in to local TV channels broadcasting select matches.
Tips for Betting Success
If you're new to sports betting or looking to refine your strategy, here are some tips to help you get started:
- Research Players: Understand player form and head-to-head records before placing bets.
- Budget Wisely: Set a budget and stick to it to avoid overspending.
- Diversify Bets: Spread your bets across different matches to minimize risk.
- Follow Trends: Stay updated with the latest news and trends that could impact match outcomes.
Fan Engagement Activities
masonchou/masonchou.github.io<|file_sep|>/_posts/2020-07-06-boosting-via-gibbs-sampling.md
---
layout: post
title: "Boosting via Gibbs Sampling"
tags: [boosting]
---
In this post we will show that AdaBoost can be obtained by Gibbs sampling.
# AdaBoost
AdaBoost (Adaptive Boosting) is one of the most popular boosting algorithms.
It was proposed by Freund & Schapire in [1997](https://link.springer.com/article/10.1023/A:1007315023581).
Let ${(x_1,y_1),dots,(x_n,y_n)}$ be training data,
where $y_iin{-1,+1}$.
We assume that $x_iinmathbb{R}^d$.
We use $H$ as hypothesis class.
The goal is to find $F(x)=sum_{t=1}^T alpha_t h_t(x)$ such that $F(x)$ classifies all training samples correctly.
AdaBoost finds $F$ iteratively.
At iteration $t$, AdaBoost finds $h_t$ such that
$$
min_{hin H} sum_{i=1}^n w_{t,i}mathbf{I}{y_ineq h(x_i)},
$$
where $mathbf{I}$ is indicator function,
and $w_{t,i}$ is weight assigned to sample $(x_i,y_i)$ at iteration $t$.
The algorithm starts with uniform weights,
and increases weight of misclassified samples.
At iteration $t+1$, weight assigned to sample $(x_i,y_i)$ is updated as
$$
w_{t+1,i}=frac{w_{t,i}}{Z_t}exp(-alpha_t y_i h_t(x_i)),
$$
where $Z_t=sum_{i=1}^n w_{t,i}exp(-alpha_t y_i h_t(x_i))$.
# Gibbs Sampling
Gibbs sampling is one of Markov Chain Monte Carlo (MCMC) methods.
It can be used to sample from joint distribution by sequentially sampling from conditional distributions.
Let $X=(X_1,dots,X_n)$ be random variables,
and let $f(X)$ be its joint probability density function.
Gibbs sampling generates samples by iteratively sampling from conditional distributions.
Let $(x_1^{(0)},dots,x_n^{(0)})$ be an initial sample.
At step $t$, Gibbs sampling generates $(x_1^{(t+1)},dots,x_n^{(t+1)})$ as follows:
$$
x_1^{(t+1)} sim f(cdot|x_2^{(t)},dots,x_n^{(t)}),\
x_2^{(t+1)} sim f(cdot|x_1^{(t+1)},x_3^{(t)},dots,x_n^{(t)}),\
dots\
x_n^{(t+1)} sim f(cdot|x_1^{(t+1)},dots,x_{n-1}^{(t+1)},x_n^{(t)}).
$$
# Boosting via Gibbs Sampling
Let $H={h_1,dots,h_M}$ be hypothesis class,
and let $alpha=(alpha_1,dots,alpha_M)$ be coefficient vector.
Let $beta=(beta_1,dots,beta_M)$ be a parameter vector such that $beta_m=exp(-alpha_m)$.
We use $hat{beta}_m=hat{beta}_m(w)$ as estimate of $beta_m$ at weight vector $w$.
Define distribution $pi(cdot|h,w)=P(H=h|W=w)$ as follows:
$$
P(h_m=1|h,w)=
begin{cases}
dfrac{w_m}{Z(w)} & m=0\
dfrac{w_m hat{beta}_m(w)}{Z(w)} & m=1,dots,M\
0 & text{otherwise}
end{cases}
$$
where $Z(w)=w_0+sum_{m=1}^M w_m hat{beta}_m(w)$.
Let $mathcal{T}$ be set of all possible training data labels,
and let $pi(cdot|w)=P(T|mathcal{T},W=w)$ be distribution over $mathcal{T}$ defined by:
$$
P(T=t|W=w)=frac{prod_{i=1}^n w_{i,t(i)}}{sum_{sinmathcal{T}} prod_{i=1}^n w_{i,s(i)}}.
$$
Let $pi(cdot|T,w)=P(H|mathcal{T},T,W=w)$ be distribution over hypotheses defined by:
$$
P(H=h|T=t,W=w)=prod_{i=1}^n P(h(t(i))=y_i|x_i,W=w).
$$
Now we are ready to prove that AdaBoost can be obtained by Gibbs sampling.
## Theorem
Let $(T,H,W)$ be random variables such that $W$ has uniform distribution on simplex,
and such that $(T,H|W=w)sim pi(cdot|cdot,w)$.
Let $(T^{(0)},H^{(0)})$ be random variables such that $(T^{(0)},H^{(0)})=(T,H)|W=w_0$ where $w_0=(n^{-d},n^{-d},dots,n^{-d})$.
If we generate $(T^{(k)},H^{(k)})$ using Gibbs sampling as follows:
$$
T^{(k+1)} sim P(cdot|h=T^{(k)},W=w_k),\
H^{(k+1)} sim P(cdot|T=T^{(k+1)},W=w_k),
$$
where $w_k=(w_k)_m=prod_{l=0}^k hat{beta}_{m,T_l(l)}$,
then we have:
$$
(E[H_m(T)]-hat{beta}_m(W))w_k=o(n^{-d}) quad m=0,dots,M,
$$
where expectation is taken with respect to $(T,H,W)sim pi(cdot|cdot,w_k)$.
## Proof
Let us define conditional distribution $tilde{pi}(A|cdot,w)=(P(A|cdot,W=w))/P(A|cdot,W=w)$.
### Lemma A
For any function $f(H,T,W)$ we have:
$$E[f(H,T,W)|H=h,T=t,W]=E[f(h,T,W)|T=t,W]=E[f(h,t,W)|W].$$
#### Proof
By definition we have:
$$E[f(H,T,W)|H=h,T=t,W]=E[f(h,T,W)|W],$$
and by definition we have:
$$E[f(h,T,W)|T=t,W]=E[f(h,t,W)|W].$$
### Lemma B
For any function $f(H,T,w)$ we have:
$$E[f(H,T,w_k)|H=h,T=t]=f(h,t,w_k).$$
#### Proof
By definition we have:
$$E[f(H,T,w_k)|H=h,T=t]=f(h,t,w_k).$$
### Lemma C
For any function $f(T,h,w_k)$ we have:
$$E[f(T,h,w_k)|T=t]=f(t,h,w_k).$$
#### Proof
By definition we have:
$$E[f(T,h,w_k)|T=t]=f(t,h,w_k).$$
### Lemma D
For any function $f(T,h,w_k)$ we have:
$$E[f(T,h,w_k)|H=h]=E[f(T,h,w_k)].$$
#### Proof
By Lemma B we have:
$$E[f(T,h,w_k)|H=h]=E[E[f(T,h,w_k)|H=h,T]|H=h] = E[E[f(h,T,w_k)|T]|H=h]. $$
By Lemma A we have:
$$E[E[f(h,T,w_k)|T]|H=h]= E[E[f(h,t,w_k)]|H=h] = E[f(h,t,w_k)]. $$
By Lemma C we have:
$$E[f(h,t,w_k)]= E[E[f(t,h,w_k)]]. $$
By Lemma B we have:
$$ E[E[f(t,h,w_k)]]= E[f(T,h,w_k)]. $$
Therefore:
$$ E[f(T,h,w_k)|H=h]= E[f(T,h,w_k)]. $$
### Lemma E
For any function $f(T,h,k)$ we have:
$lim_{krightarrow infty} P(f(T,H,k)>c | T=T(k), H=H(k)) = P(f(T,H)>c | T=T(k), H=H(k)).$
#### Proof
Consider random variables $(X,Y,Z)=(T,H,k)$. By definition we have:
$lim_{krightarrow infty} P(f(X,Y,Z)>c | X=X(k), Y=Y(k)) = P(f(X,Y)>c | X=X(k), Y=Y(k)).$
Therefore it is enough to show:
$lim_{krightarrow infty} P(Z=k | X=X(k), Y=Y(k)) = P(Z=infty | X=X(k), Y=Y(k)).$
Since $(X,Y,Z)=(T,H,K)=(K,K,K)$ with probability one then:
$lim_{krightarrow infty} P(Z=k | X=X(k), Y=Y(k)) = P(Z=infty | X=X(k), Y=Y(k)).$
### Lemma F
For any function f(H,t,k) we have:
$lim_{k->infinty} E[tilde{pi}(A|h=T(k), t=T(k)) f(H,t,k)] = E[tilde{pi}(A|h=T(k), t=T(k)) f(H,t)].$
#### Proof
Since:
$tilde{pi}(A|h=T(k), t=T(k))=frac{pi(A|h=T(k), t=T(k))}{P(A|h=T(k), t=T(k))},$
then:
$tilde{pi}(A|h=T(k), t=T(k)) f(H,t,k)=f(H,t,k) frac{pi(A|h=T(k), t=T(k))}{P(A|h=T(k), t=T(k))},$
so:
$lim_{k->infinty} E[tilde{pi}(A|h=T(k), t=T(k)) f(H,t,k)]=$
$lim_{k->infinty} E[ f(H,t,k) frac{pi(A|h=T(k), t=T(k))}{P(A|h=T(k), t=T(k))] ]=$
$lim_{k->infinty} E[ f(H,t,k) ] E[ frac{pi(A|h=T(k), t=T(k))}{P(A|h=T(k), t=T(k))] | H=h(t(K)), T=t(K)]=$
$lim_{k->infinty} E[ f(H,t,k) ] =$
$E[ lim_{k->infinty} f(H,t,k) ]=$
$E[ f(H,t) ].$
The second equality holds by Lemma D,
the third equality holds because:
$(H(t(K)), T(K)) = (K,K),$
the fourth equality holds because:
$f(H(t(K)), T(K), K) = f(K,K,K) = f(K,K),$
the last equality holds because:
$f(K,K)geq c.$
### Lemma G
For any function g(h,k) we have:
$lim_{k->infinty} E[g(h,k)] = E[g(h)].$
#### Proof
Since:
$h(t(K))=K,$
then:
$lim_{k->infinty} g(h(t(K)), k)=g(K).$
Therefore:
$lim_{k->infinty} E[g(h,k)]=$
$lim_{k->infinty} E[g(h(t(K)), k)]=$
$E[lim_{k->infinty} g(h(t(K)), k)]=$
$E[g(K)].$
### Lemma H
For any function g(t,k) we have:
$lim_{k->infinty} E[g(t,K)] = E[g(t)].$
#### Proof
Since:
$t(K)=K,$
then:
$lim_{k->infinty} g(t(K)), k)=g(K).$
Therefore:
$lim_{k->infinty} E[g(t,K)]=$
$lim_{k->infinty} E[g(t(K)), k]=$
$E[lim_{k->infinty} g(t(K)), k]=$
$E[g(K)].$
Now let us prove Theorem.
By Lemma F we have:
$(EW)^{-d}sum_w w^d(E[H_m(T)]-