# Independence of a certain Linear combination of random variables

Suppose $X$ and $Y$ are two independent standard normal random variables, and $\theta$ is a constant. Define $U = \cos(\theta)X + \sin(\theta)Y$ and $V = -\sin(\theta)X + \cos(\theta)Y$. Show that $U$ and $V$ are independent standard normal random variables.

I've shown that $U$ and $V$ are standard normal random variables, but I don't know really where to go on showing independence. All I'm familiar with here is the definition that $U$ and $V$ are independent if $\mathbb{P}(\{U \le a, V \le b\}) = \mathbb{P}(\{U \le a\}) \mathbb{P}(\{V \le b\})$. I only want something to get me started, not a full solution.

What I've tried so far is noting that $X = U\cos(\theta) - V\sin(\theta)$ and $Y = U \sin(\theta) + V \cos(\theta)$ and then trying to solve the system:

$\cos(\theta) X + \sin(\theta) Y \le a$ and

$-\sin(\theta) X + \cos(\theta)Y \le b$

in terms of $X$ and $Y$ to try and apply the independence of $X$ and $Y$. I'm pretty sure that's not going to go anywhere because it doesn't really utilize much about the definition of $U$ and $V$.

I've also recognized that formally $\frac{dU}{d\theta} = V$, but I can't really see any way that helps with showing independence.

Any suggestions on where to start?

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The joint density of $X$ and $Y$ has circular symmetry about the origin; in fact, the only random variables that are independent and have joint density that has circular symmetry are (independent) normal random variables with equal variance. Now show that your transformation $(X,Y) \to (U,V)$ is a rotation of axes for the joint density, and so still has circular symmetry (and hence $U$ and $V$ are also independent normal random variables.

Alternatively, there is a standard formulation involving Jacobians that allows you to compute $f_{U,V}$ in terms of $f_{X,Y}$. Use this and conclude that $f_{U,V}(u,v) = f_U(u)f_V(v)$ for all $u, v$ and so $U$ and $V$ are independent.

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Thank you. After locating the relevant theorem, this was very straightforward. –  anonymous Feb 7 '13 at 0:42

Let's compute the $COV$ of $U$ and $V$.

So, $COV(U,V)=E[(U-E[U])(V-E[V])]=E(UV)$

You should be able to prove $E(UV)=E[(XY(cos^{2}\theta-sin^{2}\theta))-(sin\theta cos\theta(Y^2-X^2)]=0$ and hence $U$ and $V$ are independent.

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It is also necessary to show that $U$ and $V$ are jointly normal because, in general, uncorrelated random variables are not necessarily independent. –  Dilip Sarwate Feb 7 '13 at 0:21
@DilipSarwate Linear combinations of independent normal random variables are jointly normal anyway. –  jay-sun Feb 7 '13 at 0:26
Yes, but you need to say something to that effect explicitly. The OP had only proved that $U$ and $V$ were (marginally) normal, not that they were jointly normal (and was possibly quite unaware of the fact that $U$ and $V$ are jointly normal). Incidentally, linear combinations of jointly normal (not necessarily independent) random variables are jointly normal variables/ –  Dilip Sarwate Feb 7 '13 at 2:50