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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 $Y$.

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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.

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