I have a circularly symmetric Gaussian random variable, $X = X_r+jX_i$, where $X\sim \mathcal{CN}(0, \sigma^2)$. I need to prove that $X_r$ and $X_i$ are i.i.d. random variables. Since the distribution of $X$ and $e^{j\phi}X$ are same ($\phi \in [-\pi, \pi)$), I proceed as follows:
\begin{align}
P(X\leq a+jb) &= P(X_r+jX_i \leq a+jb) \\
&= P(X_r \leq a, X_i \leq b) \\
P(e^{j\phi}X \leq a+j b) &= P((\cos\phi+ j \sin\phi)(X_r+j X_i) \leq a+j b) \\
&=P(X_r \cos\phi-X_i \sin\phi \leq a, X_r \sin\phi+X_i \cos\phi \leq b)
\end{align}
My idea was to equate the two and then prove $X_r$ and $X_i$ were identically distributed. But I do not know how to proceed. Any help would be appreciated, thanks!
I guess the splitting of inequality might be wrong, I'm not sure if that is allowed. If there is any other way of proving this result, I'm open to that as well. Thanks
Best Answer
For a circularly symmetric random variable $X$, $X$ and $e^{j\phi}X$ have same distribution. Using this the following can be proved:
\begin{align} \mathbb{E}X &= \mathbb{E}e^{j\phi}X \implies \mathbb{E}X = 0 \\ \mathbb{E}XX^T &= \mathbb{E}e^{j\phi}Xe^{j\phi}X^T \implies \mathbb{E}XX^T = 0 \end{align}
Now consider the circular Gaussian random variable given by $Z = X + jY$. Then,
\begin{align} \mathbb{E}Z &= 0 \implies \mathbb{E}X = \mathbb{E}Y = 0 \\ \mathbb{E}ZZ^T &= 0 \implies \mathbb{E}Z^2 = 0 \implies var(X) = var(Y), cov(X, Y) = 0 \end{align} Since $\mathbb{E}XY = 0$, the Gaussian random variables are uncorrelated, hence independent. Further, the mean and variances of these random variables are same, therefore they are identically distributed, completing the proof.