$X$ and $Y$ are two independent random variables, $X$ is a $Bin(2, 0.25)$ and $Y$ is a $Bin(3,0.25)$. Let $Z = X + Y$, $T = X - Y$. Find the $COV(Z,T)$. Are $Z$ and $T$ independent?
I calculate the covariance but even if it's equal to $0$ I cannot say that $Z$ and $T$ are independent, how do I solve this problem?
You can use the properties of $\def\Cov{\operatorname{Cov}}\def\V{\operatorname{Var}}\Cov$:
$$\begin{align}\Cov(Z,T)&=\Cov(X+Y,X-Y)\\&=\Cov(X,X)+\Cov(X,Y)-\Cov(X,Y)-\Cov(Y,Y)\\&=\V(X)-\V(Y)\end{align}$$
Since $\V(\mathcal{Bin}(n,p))=np(1-p)$ we get $$\begin{align}\Cov(Z,T)&=2\cdot 0.25 \cdot 0.75 - 3 \cdot 0.25 \cdot 0.75 \\&= -0.1875\end{align}$$
Clearly, they are dependent.
Intuition: If $T=0$ then $X=Y$ which means that $Z=X+Y$ is even. Without knowledge about $T$, $Z$ can also be odd. So knowing the value of $T$ can somehow change the distribution of $T$.
Intuition (2): If $Z=5$ then $X=2, Y=3$ and $T=-1$. Knowing the value of $Z$ in this case determines the value of $T$.