TAOCP 3.4.1 Exercise 15

Let $S=X_1+X_2.$ Since $X_1$ and $X_2$ are independent, the distribution function of $S$ is Condition on the value of $X_2$.

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Section 3.4.1: Numerical Distributions

Exercise 15. [HM21] If $X_1$ and $X_2$ are independent random variables with the respective distributions $F_1(x)$ and $F_2(x)$, and with densities $f_1(x) = F_1'(x)$, $f_2(x) = F_2'(x)$, what are the distribution and density functions of the quantity $X_1 + X_2$?

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Solution

Let

$S=X_1+X_2.$

Since $X_1$ and $X_2$ are independent, the distribution function of $S$ is

=\Pr{X_1+X_2\le x}.$$ Condition on the value of $X_2$. For each fixed $t$, $$\Pr{X_1+X_2\le x\mid X_2=t} =\Pr{X_1\le x-t} =F_1(x-t).$$ Therefore $$F_S(x) =\int_{-\infty}^{\infty} F_1(x-t),dF_2(t).$$ Since $F_2'(t)=f_2(t)$, $$F_S(x) =\int_{-\infty}^{\infty} F_1(x-t)f_2(t),dt.

This is the distribution function of $X_1+X_2$.

To obtain the density, differentiate with respect to $x$:

$$ f_S(x) =\frac{d}{dx}F_S(x) =\int_{-\infty}^{\infty} \frac{d}{dx}F_1(x-t),f_2(t),dt. $$

Since $F_1'(u)=f_1(u)$,

$$ f_S(x) =\int_{-\infty}^{\infty} f_1(x-t)f_2(t),dt. $$

Equivalently, replacing $t$ by $x-u$,

$$ f_S(x) =\int_{-\infty}^{\infty} f_1(u)f_2(x-u),du. $$

Thus the density of the sum is the convolution of the two densities.

Hence

$$ \boxed{ F_{X_1+X_2}(x) = \int_{-\infty}^{\infty} F_1(x-t),f_2(t),dt } $$

and

$$ \boxed{ f_{X_1+X_2}(x) = \int_{-\infty}^{\infty} f_1(x-t)f_2(t),dt = \int_{-\infty}^{\infty} f_1(t)f_2(x-t),dt. } $$

This completes the proof.

∎