Let $(X\_n)$ and $(Y\_n)$ be two sequences of random variables defined on the same probability space such that the variance of all components $X\_n$, $Y\_n$ is finite and the sequence of variances of $X\_n+Y\_n$ is non-zero and bounded. Does this imply that the variance sequences of $X\_n$ and $Y\_n$ are bounded as well?


In formulas, let $a\_n^2$ and $b\_n^2$ be the variance of $X\_n$ and $Y\_n$, respectively, and let $c\_n$ be the covariance of $X\_n$ and $Y\_n$. Then the assumption from above translates into


$$0 < \operatorname{Var} (X\_n+Y\_n) = a\_n^2 + b\_n^2 + 2c\_n \leq M$$


for some $M \geq 0$. From a purely analytical point of view, one could take $a\_n^2=b\_n^2=n+M/2$ and $c\_n=-n$. Then $(X\_n)$ and $(Y\_n)$ would have unbounded variances but the variance of $(X\_n+Y\_n)$ would be bounded. Do random variables of this form exist?



If $(X_n+Y_n)$ has bounded variance, is the same true for $(X_n)$ and $(Y_n)$?
They do. Simply choose, say, $X\_n$ a Bernoulli random variable taking values $n$ or $-n$ with probability $\frac 12$ for each. Notice that $Var(X\_n) = n^2$. Define $Y\_n = - X\_n$. You have $Var(X\_n + Y\_n) = 0$ and, to answer your last interrogation, $Cov(X\_n,Y\_n) = - Var(X\_n) = -n^2$.


**Edit** : Let now $Z$ be some random variable, as nice as you like (say, another symmetric Bernoulli, but on $\{-1,1\}$ this time) and independant from all the $X\_n$. Choose $Y\_n = - X\_n + Z$. Then $Var(X\_n + Y\_n) \equiv 1$ and $Var(X\_n) \sim Var(Y\_n) \sim n^2$.


