# If $A$ is an independent $RV$, and Var$[A+A]$ $=$Var$[A]$ + Var$[A] + 2$Cov$[A,A]$. Then Cov$[A,A]$ = Var$[A] = 0$, is this a flaw in logic?

If $$A$$ is an independent $$RV$$. Then we know that
Var$$[A+A] =$$Var$$[A] +$$Var$$[A]$$

This also means that Cov$$[A,A] = 0$$ and likewise Cov$$[A,A] =$$Var$$[A] = 0$$.

So Var$$[A+A] = 0 + 0 + 0$$.

This does not seem right to me, can someone point out my flaw in logic.

By saying $$A$$ is an independent random variable if you mean that $$A$$ is independent of itself, then all the equalities you have written are correct. In this case $$A$$ is necessarily a constant random variable so its variance and covariance with itself are both $$0$$.