Bound for covariance using the covariance properties

Question

For $a \in \mathbb R^+$ and random variables $X$ and $Y$, are these two inequalities correct?

$$2\operatorname{cov}(X,aY) \leq \operatorname{var}(X)+\operatorname{var}(aY) = \operatorname{var}(X)+a^2\operatorname{var}(Y) \tag{1}$$ $$2\operatorname{cov}(X,aY)=a \times 2\operatorname{cov}(X,Y) \leq a(\operatorname{var}(X)+\operatorname{var}(Y)) \tag{2}$$ Since in my case $0<a<<1$ and $\operatorname{var}(X) \approx\operatorname{var}(Y)$, I guess (2) is clearly better (tighter). I just wanted confirmation, thanks.

Answer 1

Let $X_1=X-E(X)$ and $Y_1=Y-E(Y)$. Then the inequality $0 \le (X_1-Y_1)^2=X_1^2+Y_1^2-2X_1Y_1$ implies that $$\text{Cov}(X,Y)=E(X_1 Y_1) \le \frac12 E(X_1^2+Y_1^2)=\frac12 \bigl(\text{Var}(X)+\text{Var}(Y)\bigr)\,.$$ Both inequalities in the question follow, and can be improved by a factor of 2.