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? 🙂
Best Answer
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.