I have a random variable $X$ which is a complex Gaussian with mean equal to 0 and variance equal $\sigma^2$. I want to know the probability of $|X|$ exceeding a bound $B$. Now, the complex Gaussian with variance $\sigma^2$ has the same distribution as a 2-D Gaussian with variance $\sigma^2/2$. The problem I have is the following: why is the following expression true?
$$P[|X|>B] = \exp(-B^2/\sigma^2)$$
To me it seems unintuitive, as the exponential is the probability distribution and not the cumulative distribution. Any insights are welcome. It is mentioned that this is a heuristic bound if that helps.
Best Answer
I'm going to assume that when you say complex gussian with mean $0$ and variance $\sigma^2/2$, you mean a 2D multivariate normal distribution with mean vector $\bf0$ and covariance matrix $\frac{\sigma^2}{2}I$. Then $$R^2 = X_1^2 + X_2^2 = \frac{\sigma^2}{2} {\chi^2_2}$$ Where $\chi^2_2$ is a random variable with Chi-square distribution. Now for the question $$P(|X| > B) = P\bigg(\chi^2_2 > \frac{2B^2}{\sigma^2}\bigg) = \exp\bigg(-\frac{B^2}{\sigma^2}\bigg)$$