Consider two sums of independent random variables, $S_1 = \sum_{i=1}^N X_i$ and $S_2 = \sum_{i=1}^N Y_i$. Both $X_i$ and $Y_i$ take only values $0$ or $1$ for all $i$ and they are not equally distributed. What is the shortest and most elegant way to prove that there exists $K>0$ such that for all $N$ the following holds? $$P( S_2 - S_1 \geq E[S_2] - E[S_1]) \geq K $$
Can one apply the Central Limit Theorem in this case?
If $(X_i)$ is i.i.d., $(Y_i)$ is i.i.d. and $(X_i)$ and $(Y_i)$ are independent, this follows from the central limit theorem applied to the i.i.d. sequence $(Z_i)$ defined by $$ Z_i=X_i-Y_i-E(X_1)+E(Y_1). $$ To wit, considering the events $$ A_N=[S_2-S_1\geqslant E(S_2)-E(S_1)], $$ one gets $$ A_N=[U_N\geqslant0]=\left[\frac{U_N}{\sqrt{N}}\geqslant0\right],\qquad U_N=Z_1+\cdots+Z_N.$$ Introducing some standard normal random variable $U$, the central limit theorem shows that, when $N\to\infty$, $$ P(A_N)\to P(U\geqslant0)=\tfrac12, $$ hence $P(A_N\geqslant0)\geqslant\frac13$ for every $N$ large enough, say $N\geqslant N_0$. On the other hand, $P(A_N)\gt0$ for every $N\leqslant N_0$ hence $$ \inf\limits_{N\geqslant1}P(A_N)\gt0. $$