From this question here, I learned that the Cumulative Distribution Function (CDF) of $Z=X+Y$ is: \begin{eqnarray*} F_Z \left( z \right) & = & \int F_X \left( z - y \right) dF_Y \left( y \right)\\ & = & \int F_Y \left( z - x \right) dF_X \left( x \right) \end{eqnarray*}

And Probability Density Function (PDF) of $Z=X+Y$ is:

\begin{eqnarray*} f_Z \left( z \right) & = & \int f_X \left( z - y \right) f_Y \left( y \right)dy\\ & = & \int f_Y \left( z - x \right) f_X \left( x \right)dx \end{eqnarray*}

How about PDF and CDF of sum of weighted independent random variables $Z=aX+bY$ where $a$ and $b$ are positive integers? I tried finding in this site but no luck. Thanks


1 Answer 1


\begin{align*} F_Z(z)&=Pr(aX+bY<z)=\int Pr(aX+by<z)f_Y(y)dy\\ &=\int F_X\left(\frac{z-by}{a}\right)f_Y(y)dy\\ f_Z(z)&=\frac{1}{|a|}\int f_X\left(\frac{z-by}{a}\right)f_Y(y)dy \end{align*}

  • 1
    $\begingroup$ thank you for the detailed answer! i will check them numerically soon $\endgroup$ Commented Mar 17, 2015 at 11:49

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