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 $$ f_{z}(z) = \int_{-\infty}^{\infty}f_X(z-y)f_Y(y)dy = \int_{-\infty}^{\infty}f_X(x)f_Y(z-x)dx .$$

  • $\begingroup$ 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 : $P(X+Y \le z) = \sum_i P(X = x_i) P(Y \le z-x_i)$ $\endgroup$ – reuns Sep 8 '16 at 18:46
  • 2
    $\begingroup$ 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. $\endgroup$ – Did Sep 8 '16 at 19:29
  • $\begingroup$ 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. $\endgroup$ – Randerson Sep 8 '16 at 22:24
  • $\begingroup$ 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 @.) $\endgroup$ – Did Sep 9 '16 at 6:21
  • $\begingroup$ I am sorry... I am messing everything. Anyway, thanks. $\endgroup$ – Randerson Sep 9 '16 at 12:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.