Help needed in understanding the convolution of two distributions

Question

I'm reading Introduction to Probability Models by Sheldon Ross, 12th edition. On page 57, it says:

It is often important to be able to calculate the distribution of $X + Y$ from the distributions of $X$ and $Y$ when $X$ and $Y$ are independent. Suppose first that $X$ and $Y$ are continuous, $X$ having probability density $f$ and $Y$ having probability density $g$. Then, letting $F_{X+Y}(a)$ be the cumulative distribution function of $X + Y$, we have

$$ F_{X+Y}(a) = P(X + Y \leq a) $$ $$ = \int_{x+y \leq a} f(x) g(y) \, dx \, dy $$ $$ = \int_{-\infty}^{\infty} f(x) \left( \int_{-\infty}^{a - x} g(y) \, dy \right) dx $$ $$ = \int_{-\infty}^{\infty} F_Y(a - x) f(x) \, dx $$

The cumulative distribution function $F_{X+Y}$ is called the convolution of the distributions $F_X$ and $F_Y$ (the cumulative distribution functions of $X$ and $Y$, respectively).

---

But, we must note that, $X+Y$ is a random variable. I think $X+Y$ is not necessarily continuous. It may be discrete as well. Let $Z=X+Y.$

If $Z$ is continuous then, let's assume that $f_{Z}$ is the probability density function. Under this assumption, we must have, $P\{Z\leq a\}=P\{X+Y\leq a\}=\int_{-\infty}^af_Z(z)dz.$

If $Z$ is discrete then, let's assume that $F_Z$ and $p_z$ is the cumulative distribution function and the probability mass functiom of $Z$ respectively. Under this assumption, we have, $P\{Z\leq a\}=F_Z(a)=\sum_{x\leq a:p(x)>0}p(x).$

But I don't get how do they write the equation,

$$ F_{X+Y}(a) = P(X + Y \leq a) $$ $$ = \int_{x+y \leq a} f(x) g(y) \, dx \, dy $$ $$ = \int_{-\infty}^{\infty} f(x) \left( \int_{-\infty}^{a - x} g(y) \, dy \right) dx $$ $$ = \int_{-\infty}^{\infty} F_Y(a - x) f(x) \, dx $$

Can you please help me understand, how did they write the above equation?

Answer 1

The derivation in the book is correct, and so is the final formula. But when the author wrote
$F_{X+Y}(a) = \int_{x+y\le a} f(x)g(y)dxdy$
it should really have been be written as
$F_{X+Y}(a) = \int_{x=-\infty}^{x=+\infty}\int_{y=-\infty}^{y=a-x}f(x)g(y)dxdy$
Authors should not use a single integral, when really we are dealing with a double integral. But...You then take $f(x)$ out of the 2nd integral to obtain
$F_{X+Y}(a) = \int_{x=-\infty}^{x=+\infty}f(x)\bigg(\int_{y=-\infty}^{y=a-x}g(y)dy\bigg)dx$
Now the 2nd integral is $F_Y(a-x)$, i.e. the CDF of Y at $y=a-x$ (it is confusing to use $F$ for all CDF's, when one uses $f$ and $g$ for the PDF's; one should probably have used $G(a-x)$, but...). You now get
$F_{X+Y}(a) = \int_{x=-\infty}^{x=+\infty}f(x)F_Y(a-x)dx$

BUT, and kind of a big but, I take exception to the next sentence;
"The cumulative distribution function $F_{X+Y}$ is called the convolution of the distributions $F_X$ and $F_Y$ (the cumulative distribution functions of X and Y, respectively)".
That is just plain wrong. The CDF of the sum is not the convolution of the CDF's (which is what the author wrote). It is the convolution of the CDF of one distribution by the PDF of another. It is the PDF of the sum which is the convolution of the PDF's of the 2 distributions. If we use $*$ to denote the convolution operator, we have
$f_{X+Y}(a)=(f_X*f_G)(a)$ and $F_{X+Y}(a)=(f*F_Y)(a)=(F_X*g)(a)$

So the formula is correct, I hope the explanations clarified it, but the last sentence is absolutely wrong.