Contradiction with complex gaussians...

Question

So, I am computing something seemingly simple involving complex gaussians and constants, but I am getting a big contradiction in my calculations.

The setup:

• Let $C$ be a complex constant, that is, $C = c_r + jc_i$.
• Let $G$ be a complex gaussian variable, $G = g_r + jg_i$, where $g_r$ and $g_i$ are both uncorrelated, and where each are $\sim\mathcal{N}(0,\sigma^2)$.

I am computing $z = |C + G|^2$.

The problem: Now, before I go on, it is obvious that the variable $z$, must always be greater than or equal to $0$, owning to the $| \cdot |^2$ operation. However when I open up and compute $z$, I get an expression that seems like it CAN be less than $0$.

Opening up $z$, I get

$$ z = (c_r^2 + c_i^2) + 2\Big[c_rg_r + c_ig_i \Big] + (g_r^2 + g_i^2) $$

The first term is a constant, and will always be greater than or equal to zero. The last term is has a gamma distribution, and by definition, will also always be greater than or equal to zero. However, the middle term is simply a summation of two gaussians, but this means that there is a finite probability that they take on a value of less than zero, meaning that $z$ can also be less than zero!

But $z$ can never be less than zero. This is the contradiction... I am not sure where I am making a mistake in my reasoning...

Thank you.

Answer 1

The mistake is the fact that if you have some dependent expressions, and some are positive and some can be negative, you conclude that the result can be negative.

Sometimes the dependency between the expressions forces the sum to always be positive.

To understand the issue better look at the following different simpler example: Pick a number $x >2$ and another number $y \in (-1,0)$. Now, $x$ is positive and $y$ is negative, can you make $x+y$ negative?

In you example, while the middle term can be negative, it follows immediately from the AM-GM inequality that the smallest value it can take is

$$-\frac{(c_r^2 + c_i^2) + (g_r^2 + g_i^2)}{2}$$

Because of this, $z$ will always be greater or equal than 0.

Answer 2

One thing was initially left unclear: You wrote $c_i+jg_i$ without telling us that $c_i$ is constant and $g_i$ is a random variable.

It is true that $2c_i g_i$ may be negative. But that does not mean $c_i^2+2c_i g_i + g_i^2$ can be negative. In fact $c_i^2+2c_i g_i + g_i^2=(c_i+g_i)^2$, so this cannot be negative.

It is not correct to say that $c_i^2+g_i^2$ is a gamma distribution. Rather $c_i^2+g_i^2$ has a gamma distribution.

How would one know that $4x^2-12x+25$ can never be negative, and in fact can never be $0$, when $-12x$ can be negative? The most informative answer is by completing the square: $$ 4x^2-12x+25 = 4(x^2-3x)+25 = 4\left(x^2-3x + \frac 9 4\right) + 25 - 9 = 4\left(x - \frac 3 2\right)^2 + 16. $$ This is a square plus $16$, so it's always positive no matter what number $x$ is.

Your situation is simpler than that. Let's complete the square: $$ c_i^2 + 2c_i g_i + \text{what} = \text{a square?} $$ In place of "what" in this case we'd but $g_i^2$, and we see the the square is already complete.