Let $X_n$ $Y_n$ be a sequence of random variables such that for $C > 0$ \begin{align*} 0 \leq X_n + Y_n \quad \text{and} \quad -C \leq X_n \end{align*} Suppose that for positive sequences $a_n,\, b_n \, c_n$ which converge to zero \begin{align*} E(X_n) & \leq a_n \\ E(|Y_n| ) &\leq b_n \\ Y_n & = \mathcal{O}_P(c_n) \end{align*} If $c_n = o(b_n)$ is it true that $$X_n + Y_n = \mathcal{O}_P(a_n + c_n) ?$$
By Markov Inequality we get $X_n +Y_n = \mathcal{O}_P(a_n + b_n)$ but I belive there might be room for some improvement.
The notation $ Y_n = \mathcal{O}_P(c_n)$ is Big O in Probability notation. I.e
Given $\varepsilon >0$ there exists a constant $C >0$ and $n_0$ such that $P(|Y_n| > C c_n) \leq \varepsilon$ for all $n \geq n_0$.
Set $(X_n,Y_n)=(-1,1)$ with probability $b_n,$ and $(X_n,Y_n)=(b_n/(1-b_n),0)$ with probability $1-b_n.$ Note $X_n+Y_n$ is $b_n/(1-b_n)\sim b_n$ with probability tending to $1.$
For example set $a_n=b_n=1/n$ and $c_n=1/n^2.$ Then $X_n+Y_n$ is not $\mathcal O_P(1/n^2).$