Probability as integral of itself w.r.t. the distribution function I'm having trouble understanding the following.
$(T_n)_{n\in\mathbb{N}}$ is a sequence of random variables. Then:
$$
P(T_n \leq t)=\int_0^t P(y+(T_n-T_1) \leq t) \text{d} F_{T_1} (y)
$$
Don't know if it matters, but $T_n=\sum_{i=1}^n W_i$ where $W_i$ are iid.
In short, what is going on here? We integrate the distribution with respect to itself, and what we get is that same distribution function?
 A: This is the law of total probability for continuous random variables. In fact,
\begin{align}
\mathbb{P}(T_n\le t)&=\mathbb{E}(\mathbb{1}_{T_n\le t})\\
&=\mathbb{E}(\mathbb{E}(\mathbb{1}_{T_n\le t}|T_1))\\
&=\int_{\mathbb{R}}\mathbb{E}(\mathbb{1}_{T_n\le t}|T_1=y)\,{\rm d}F_{T_1}(y)\\
&=\int_{\mathbb{R}}\mathbb{P}(T_n\le t|T_1=y)\,{\rm d}F_{T_1}(y).
\end{align}
The independence between $W_i$'s yields
$$
\mathbb{P}(T_n\le t|T_1=y)=\mathbb{P}\biggl(y+\sum_{i=2}^nW_i\le t\biggr)=\mathbb{P}(y+\left(T_n-T_1\right)\le t).
$$
Therefore,
$$
\mathbb{P}(T_n\le t)=\int_{\mathbb{R}}\mathbb{P}(y+\left(T_n-T_1\right)\le t)\,{\rm d}F_{T_1}(y).
$$
Finally, I suppose $W_i\ge 0$ for all $i=1,2,...,n$. In this case,
\begin{align}
T_1&\ge 0,\\
T_n-T_1&\ge 0.
\end{align}
These two facts reduce the integral domain from $\mathbb{R}$ to $\left[0,t\right]$, i.e.,
$$
\mathbb{P}(T_n\le t)=\int_0^t\mathbb{P}(y+\left(T_n-T_1\right)\le t)\,{\rm d}F_{T_1}(y).
$$
This completes the proof.
A: Simply using the definition of $T_n$ and $W_1, \ldots, W_n$ are i.i.d. $\sim F \equiv F_{T_1}$, whence $T_n - T_1 = W_2 + \cdots + W_n$ is independent of $T_1 \equiv W_1$. It follows that 
\begin{align}
P(T_n \leq t) & = P(T_n - T_1 + T_1 \leq t) = \int_{-\infty}^\infty P(T_n - T_1 + y \leq t) dF_{T_1}(y) \tag{$*$}
\end{align}
If you further require that $W_i$s are non-negative, then the integration limits would be as you posted. 
In $(*)$, we used the very useful expression in probability: if two random variables $X$ and $Y$ are independent, then for any $t \in \mathbb{R}$, 
\begin{align}
P(X + Y \leq t) = \int_{\mathbb{R}} P(X \leq t - y) dF_Y(y) = \int_{\mathbb{R}} P(Y \leq t - x) dF_X(x).
\end{align}
