X and Y are independent random variables with the same distribution X and Y are independent random variables with the same distribution. If W=X+Y and Z=X-Y why are W and Z not independent?
We give an example that may make it clear why $X-Y$ and $X+Y$ need not be independent. 

Let $X$ and $Y$ be (independent and) uniformly distributed on $[0,1]$.

Suppose that $|X-Y|$ is "large" (bigger than $0.9$). Then it is impossible for $X+Y$ to be small (say less than $\frac{1}{2}$). So information about $X-Y$ ccan yield information about $X+Y$.  

Remark: One could rewrite the above as $\Pr\left((|X-Y|\gt 0.9)\cap (X+Y\lt \frac{1}{2})\right)\ne \Pr(|X-Y|\gt 0.9)\Pr(X+Y\lt \frac{1}{2})$, but that would somewhat defeat the point of the example.
If two variables are independent, then they must have a covariance of 0.
$$
\begin{align}
Cov(W,Z) &amp;= Cov\left((X+Y),(X-Y)\right)\\
Cov\left((X+Y),(X-Y)\right) &amp;= E\left[(X+Y)(X-Y)\right] - \left[E(X+Y)\right]\left[E(X-Y)\right]\\
&amp;= E(X^2 - Y^2) - \left[E(X) + E(Y)\right]\left[E(X) - E(Y)\right]\\
&amp;= E(X^2) - E(Y^2) - E(X)^2 + E(Y)^2\\
&amp;= \left[E(X^2) - E(X)^2\right] - \left[E(Y^2) - E(Y)^2\right]\\
&amp;= Var(X) - Var(Y)
\end{align}
$$

Just because $X$ and $Y$ are independent does not mean that they have the same variance. For cases where they do not have the same variance, the covariance term is &lt;> 0, and thus the sum and difference of two independent random variables is not necessarily itself independent.