# Prove that vector has normal distribution

You are given two independent random variables: $W \sim \mathrm{Exp}(1)$, $Q \sim U([0; 2\pi ])$.
Also, $a$ is a constant, chosen from $[-\pi/2; \pi/2]$.

You build following random variables, based on $R, W, Q, a$:
$R = \sqrt{2W}$,
$U = R \cos Q$,
$V = R \sin (Q + a)$.

The task is to prove that vector $(U, V)$ has bivariate normal distribution with correlation $\rho = \sin a$.

It is almost obvious that both $U$ and $V$ have zero expectation and odd variance:
$EU = EV = 0$, ${\sigma}_{U} = {\sigma}_{V} = 1$,
so it's not very difficult to prove, that correlation of $U$ and $V$ is $\sin a$. Though, I find it hard to prove, that the vector has normal distribution, so I'd really appreciate your help.

P.S.: Don't you find it amazing, that we can construct normal distribution based on exponential and uniform ones? :)

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Search for Box-Muller method. –  Dilip Sarwate Dec 20 '11 at 20:15
You can generate from two uniform distributions with $W$ uniform on $(0,1]$ and $R=\sqrt{-2 \log_e W}$ –  Henry Dec 20 '11 at 20:22
Yep, Box-Muller transformation is very similar to the task described above. Still, Box-Muller generates independent normally distributed values (they correspond for a=0), and I am searching for a provement for any other "a" from the interval. –  user21394 Dec 20 '11 at 20:29
If you proved already that for $a=0$ the distributions of $U$ and $V$ are normal, then you can easily show that it is also the case for $a\ne 0$. For $U$ it is the exactly the same calculation (it is independent of $a$!). For $V$ you might use $\sin(Q+a)=\cos(Q)\sin(a)+\sin(Q)\cos(a)$ So $V$ is a sum of two normal variables, which are independent (because you showed that for $a=0$, and therefore also a normal variable. –  yohBS Dec 20 '11 at 21:18
Really, it is. Thanks for help! –  user21394 Dec 20 '11 at 21:43

After more careful reading of your question, I think that what you really want is a proof that $R\cos(Q)$ and $R\sin(Q+a)$ are jointly normal random variables and have the bivariate joint normal density that you are familiar with. The other stuff, means, variances, covariance etc you have been able to work out for yourself. As @yohBS points out in his comment, $$R\sin(Q+a) = R\cos(Q)\sin(a)+R\sin(Q)\cos(a)$$ is a linear combination of $R\cos(Q)$ and $R\sin(Q)$ which you know are independent and therefore jointly normal random variables.

Edited in response to Didier Piau's comments

Many people define jointly normal random variables in terms of linear functionals: $X$ is a vector of jointly normal random variables (normal vector for short) if and only if every linear functional of $X$ yields a normal random variable. Affine functionals also yield normal random variables. (Constants are honorary normal random variables with variance $0$). Thus, if $X$ is a normal vector, then $Y = a_0+\sum a_iX_i$ and $Z = b_0+\sum b_iX_i$ are normal random variables, and since $$\alpha Y + \beta Z = (\alpha a_0+\beta b_0) + \sum_i (\alpha a_i+\beta b_i)X_i$$ is a normal random variable for all $\alpha$ and $\beta$, $(Y,Z)$ is a normal vector. More generally, if $X$ is a normal vector, then $AX+B$ is a normal vector for any matrix $A$ and real vector $B$.

If $X$ is a single continuous random variable with pdf $f_X(x)$, then for $a \neq 0$, $Y = aX+b$ is a continuous random variable with pdf $f_Y(y) = \frac{1}{|a|}f_X\left(\frac{y-b}{a}\right)$. Similarly, for a random $n$-vector $X$ of jointly continuous random variables with joint pdf $f_X(x)$, invertible $n\times n$ matrix $A$ and real vector $B$, $Y = AX+B$ is a $n$-vector of jointly continuous random variables with joint pdf $$f_Y(y) = \frac{1}{|\text{det}(A)|}f_X(A^{-1}(y-B))$$ where $\text{det}(A)$, the determinant of $A$ is coming from the Jacobian of the linear transformation.

For a random $n$-vector $X$, the mean vector $E[X]$ is a real vector with $i$-th coordinate $E[X_i]$ and the $n\times n$ covariance matrix $S$ is $E[(X-E[X])(X-E[X])^T]$ whose $\{i,j\}$-th entry is $\text{cov}(X_i,X_j) = E[(X_i-E[X_i])(X_j-E[X_j])]$. Suppose now that the $X_i$ are independent standard normal random variables and so the joint pdf of the $X_i$'s is $$f_X(x) = \prod_{i=1}^n f_{X_i}(x_i) = \prod_{i=1}^n \frac{1}{\sqrt{2\pi}}\exp(-x_i^2/2) = (2\pi)^{-n/2}\exp\left(-\frac{1}{2}x^Tx\right).$$ Then, $Y = AX+B$ is a normal vector, and it has mean vector $$E[Y] = E[AX+B] = AE[X]+B = B,$$ and covariance matrix $$S_Y = E[(Y-B)(Y-B)^T]= E[AXX^TA^T] = AE[XX^T]A^T = AA^T$$ since $E[XX^T] = I$, the identity matrix. Furthermore, the joint pdf of the $Y_i$'s is \begin{align*} f_Y(y) &= \frac{1}{|\text{det}(A)|}f_X(A^{-1}(y-B))\\ &= \frac{1}{(2\pi)^{n/2}|\text{det}(A)|} \exp\left(-\frac{1}{2}(A^{-1}(y-B))^TA^{-1}(y-B)\right)\\ &=\frac{1}{(2\pi)^{n/2}|\text{det}(A)|} \exp\left(-\frac{1}{2}(y-B)^T(A^{-1})^TA^{-1}(y-B)\right)\\ &= \frac{1}{(2\pi)^{n/2}|\text{det}(A)|^{1/2}|\text{det}(A^T)|^{1/2}} \exp\left(-\frac{1}{2}(y-B)^T(AA^T)^{-1}(y-B)\right)\\ &= \frac{1}{(2\pi)^{n/2}|\text{det}(AA^T)|^{1/2}} \exp\left(-\frac{1}{2}(y-B)^T(AA^T)^{-1}(y-B)\right)\\ &= \frac{1}{(2\pi)^{n/2}|\text{det}(S_Y)|^{1/2}} \exp\left(-\frac{1}{2}(y-B)^TS_Y^{-1}(y-B)\right) \end{align*}

In this world view, $[R\cos(Q), R\sin(Q+a)]^T$ is a normal vector since it is obtained by linear transformation from $[R\cos(Q), R\sin(Q)]^T$ which is a normal vector of standard normal random variables, and so the joint pdf of $R\cos(Q)$ and $R\sin(Q+a)$ is a bivariate normal density that can be worked out from the general form above.

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About jointly normal random variables: the canonical definition is that every linear functional $\ell$ applied to the vector valued $X$ yields a normal real valued random variable $\ell(X)$. In other words, one does not consider general linear transformations (by the way, doing so would yield a circular definition). // A little further in your answer, one could think that a Delta mass at zero is the only degenerate case, but not so: every Dirac mass may occur, and indeed these all should be considered as bona fide normal distributions. –  Did Dec 26 '11 at 12:40
@DidierPiau You are right as usual. I have revised my answer and hope that it has improved in the process. –  Dilip Sarwate Dec 28 '11 at 4:53
It has. Definitely. –  Did Dec 28 '11 at 13:11