# Sum of Random Variables and Convolution

I am trying to show that the pdf of the sum of two independent random variables is the convolution of their pdf but I not succeed in this. What I am doing is: Be $X$ and $Y$ two independent random variable and be $Z$ equal to $X + Y$. Hence, $Z$ is a random variable with cumulative distribution function equals to (these first steps are also describes here)

\begin{aligned} F_Z(z) = P(\{Z \leq z\}) = P(\{X+Y \leq z\}) \end{aligned},

\begin{aligned} F_Z(z) &= \int\int_{(x,y):x+y\leq z}f_{X,Y}(x,y)dxdy, \\ &= \int_{-\infty}^{\infty}dx\int_{-\infty}^{z-x} f_{X,Y}(x,y)dy. \end{aligned}

Since $X$ and $Y$ are independent random variables $f_{X,Y}(x,y) = f_X(x)f_Y(y)$, we have \begin{aligned} F_Z(z) &= \int_{-\infty}^{\infty}dx\int_{-\infty}^{z-x} f_X(x)f_Y(y)dy,\\ &= \int_{-\infty}^{\infty}dx\int_{-\infty}^{z} f_X(x)f_Y(t-x)dt. \end{aligned}

With the cumulative distribution function of $Z$, the associated pdf can be found by taking the derivative of the function with respect $z$. Doing this we have $$\frac{d}{dz}F_z(z) = \frac{d}{dz}\Bigg\{\int_{-\infty}^{\infty}dx\int_{-\infty}^{z} f_X(x)f_Y(t-x)dt\Bigg\}.$$

And here I am stuck, because I just do not know how to perform this derivative in the manner to obtain $$f_{z}(z) = \int_{-\infty}^{\infty}f_X(z-y)f_Y(y)dy = \int_{-\infty}^{\infty}f_X(x)f_Y(z-x)dx .$$

• you have to show that $P(X+Y \le z) = \int_{-\infty}^\infty P(Y \le z-x) dF_X(x)$. If $X$ is a discrete random variable, it becomes obvious : $P(X+Y \le z) = \sum_i P(X = x_i) P(Y \le z-x_i)$ – reuns Sep 8 '16 at 18:46
• You already did nearly all the work: simply rewrite the double integral just before your "With the cumulative distribution function of Z" as $$F_Z(z) =\int_{-\infty}^{z} \left(\int_{-\infty}^{\infty} f_X(x)f_Y(t-x)dx\right) dt$$ and now the PDF of $Z$ should be apparent. – Did Sep 8 '16 at 19:29
• In this way, if I take the derivative with respect to $dz$ it is implicit that $dz=dt$, but I can not see straightforward this relation. – Randerson Sep 8 '16 at 22:24
• What? You know that $F_Z(z)=\int_{-\infty}^zg(t)dt$ for some function $g$ and you cannot deduce the derivative with respect to $z$? (Unrelated: Please use @.) – Did Sep 9 '16 at 6:21
• I am sorry... I am messing everything. Anyway, thanks. – Randerson Sep 9 '16 at 12:22