# PDF of sum of two circular symmetric complex random variables

Suppose that $$X = X_R + j X_I$$ and $$Y = Y_R + j Y_I$$ are two circular symmetric complex random variables, can we use the convolution operation to calculate the PDF of $$Z = X + Y$$, i.e., $$f_Z(u) = \int_{\mathbb{R}^2} f_X(u - v) f_Y(v) dv \,?$$ where $$f_X$$ and $$f_Y$$ are joint PDFs of $$X$$ and $$Y$$.

## 1 Answer

https://en.wikipedia.org/wiki/Convolution#Distributions

Let see if that is what you want:

Let $$(X_1, X_2)$$ and $$(Y_1, Y_2)$$ be two pairs of random variables with joint pdf $$f_{X_1, X_2}, f_{Y_1, Y_2}$$ respectively, and they are independent. Define $$(Z_1, Z_2) = (X_1 + Y_1, X_2 + Y_2)$$. Consider the joint CDF of $$(Z_1, Z_2)$$:

\begin{align} F_{Z_1, Z_2}(z_1, z_2) &= \Pr\{Z_1 \leq z_1, Z_2 \leq z_2\} \\ &= \int_{\mathbb{R}^2} \Pr\{Y_1 \leq z_1 - x_1, Y_2 \leq z_2 - x_2\}f_{X_1, X_2}(x_1, x_2)d(x_1, x_2) \end{align} Differentiate it to obtain the joint pdf of $$(Z_1, Z_2)$$: \begin{align} f_{Z_1, Z_2}(z_1, z_2) &= \frac {\partial^2} {\partial z_1 \partial z_2}F_{Z_1, Z_2}(z_1, z_2) \\ &= \frac {\partial^2} {\partial z_1 \partial z_2}\int_{\mathbb{R}^2} \Pr\{Y_1 \leq z_1 - x_1, Y_2 \leq z_2 - x_2\}f_{X_1, X_2}(x_1, x_2)d(x_1, x_2) \\ &= \int_{\mathbb{R}^2} \frac {\partial^2} {\partial z_1 \partial z_2} \Pr\{Y_1 \leq z_1 - x_1, Y_2 \leq z_2 - x_2\}f_{X_1, X_2}(x_1, x_2)d(x_1, x_2) \\ &= \int_{\mathbb{R}^2} f_{Y_1, Y_2}(z_1 - x_1, z_2 - x_2)f_{X_1, X_2}(x_1, x_2)d(x_1, x_2) \\ &= \int_{\mathbb{R}^2} f_{\mathbf{Y}}(\mathbf{z}-\mathbf{x})f_{\mathbf{X}}(\mathbf{x})d\mathbf{x} \end{align}

Not very sure if I am allowed to write the last sentence compactly for the vectors.