If $(X_n+Y_n)$ has bounded variance, is the same true for $(X_n)$ and $(Y_n)$? [closed]

Question

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?

Answer 1

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 $Var(X_n) \sim Var(Y_n) \sim n^2$.