Sums of random variables expressed as convolution If $X_1, \dots, X_n$ are a sequence of IID random variables, and $S_1, \dots, S_n$ is the sum of the first $n$ random variables, i.e. $S_1 = X_1$ and $S_n = \sum_{i = 1}^{n} X_i$.
Apparently, we can go from:
$S_n = S_{n-1} + X_n$
to the integral equation
$$P(S_n \leq t) = \int_0^t  P(S_{n-1} \leq t - x) \, dF_X(x)$$
I'm assuming (but am not sure) $F_X = P(X_n < t) = P(X_1 < t)$ because of the $X$'s being IID random variables (NOT assuming a Poisson Process).
And apparently this has something to do with convolution??
Can someone please explain whether the integral equation is correct, and how to derive it?
 A: $\newcommand{\E}{\operatorname E}$
You can do this if you know a couple of things.  One of them is
$$
\E(g(X)) = \int_{-\infty}^\infty g(x)\,dF_X(x)
$$
where $F_X(x)=P(X\le x)$.  Another is the law of total probability:
$$
\E(P(A\mid X)) = P(A).
$$
Let's apply this to the event $A=[S_n\le t]$.  We have
$$
P(S_n\le t) = P(S_{n-1}+X_n\le t) = \E(P(S_{n-1} +X_n\le t\mid X_n)).
$$
To find this, consider
$$
P(S_n\le t\mid X_n=x)=
P(S_{n-1} +X_n\le t\mid X_n=x) = P(S_{n-1}+x\le t) = P(S_{n-1}\le t-x).
$$
Hence
$$
\begin{align}
\E(P(S_n\le t\mid X_n)) & = \int_{-\infty}^\infty P(S_n\le t \mid X_n=x)\, dF_{X_n}(x) \\[10pt]
& = \int_{-\infty}^\infty P(S_n\le t-x)\,dF_{X_n}(x).
\end{align}
$$
It's a sort of convolution because you have a function of $t$ expressed as an integral with respect to $x$ of a function of $t-x$ and a function of $x$.  If the distribution of $X_n$ is absolutely continuous then $\displaystyle \int_{-\infty}^\infty(\cdots\cdots)\,dF(x)$ is the same as $\displaystyle\int_{-\infty}^\infty (\cdots\cdots)\,\cdot F'(x)\,dx$.  But the inftegral involving $dF(x)$ is valid even when $F$ is not absolutely continuous.
PS: The form with $\displaystyle\int_0^t$ instead of $\displaystyle\int_{-\infty}^\infty$ makes sense when all of the random variables are nonnegative.  That's because when either $x<0$ or $t-x<0$, the expression being integrated is $0$.
