Two days ago a friend of mine questioned me about these simple statistics problem:
Q: Take X and Y as two independet random variables with $X \sim N(0,1)$ and $Y \sim N(0,2)$. Calculate de covariance between $T=X+Y$ and $U=X-Y$.
Well, one can clearly conclude:
$$Cov(T,U)= Cov(X+Y,X-Y)= Cov(X,X) - Cov(X,Y) + Cov(Y,X) - Cov(Y,Y)=Var(X)-Var(Y) $$
But, $Var(x)=1$ and $Var(Y)=2$, therefore:
$$Cov(T,U)=-1$$
Nevertheless, we know from the Central Limit Theorem that $T \sim N(0,3)$ and $U\sim N(0,3)$. So I wonder how two variables with same distribution have a negative covariance.
Thank you very much.
There's norhing impossible about it. It just means that if you measure $T$ and $U$ seperately, each of them has a normal distribution, but when you comapre the results, when $T$ is positive, then $U$ is more likely to be negative than positive, and when $T$ is negative then $U$ is more likely to be positve than negative.
Consider a simpler case, where $T \sim N(0,1)$, $U=-T$. Then also $U \sim N(0,1)$ and $$ Cov(T,U) = -Cov(T,T) = -Var(T) =-1$$ You also have the same normal distribution and negative corelation.
Or you can consider $ X,Y\sim N(0,1)$, with $Cov(X,Y)=0$ and $T=2X\sim N(0,4)$, $U=-X+\sqrt{3}Y \sim N(0,4)$. Then you also have $$Cov(T,U) = -2Cov(X,X)+2\sqrt{3} Cov(X,Y) = -2 $$
Remember, having the same distribution doesn't mean that two variables are equal.