Constructing a Poisson process with Exponentially distributed random variables

Question

If we have iid random variables $X_1, X_2, \dots$ each having an exponential density with parameter $\lambda$, then the sum $Y_n = X_1 + \dots + X_n$ has the Gamma density $\Gamma(n,\lambda)$. We can show that: \begin{eqnarray} {\mathbb P}[Y_n \leq t < Y_{n+1}] & = & {\mathbb P}[Y_n \leq t < Y_{n} + X_{n+1}] \\ & = &\int_{0}^{t} {\mathbb P}[X_{n+1} > t - y] f_{Y_n}(y)dy \;\;\;(\mbox{Why}?)\\ & = &\int_{0}^{t} e^{-\lambda (t-y)}\frac{\lambda^n y^{n-1}}{(n-1)!}e^{-\lambda y} dy \\ &=& \frac{(\lambda t)^n}{n!}e^{-\lambda t} \end{eqnarray} The above can be used to show that a counting process consisting of the $Y_n$ is a Poisson process with rate $\lambda$. However, it's not clear to be why the second line holds true? Why does: $$ {\mathbb P}[Y_n \leq t < Y_{n} + X_{n+1}] = \int_{0}^{t} {\mathbb P}[X_{n+1} > t - y] f_{Y_n}(y)dy $$

Answer 1

$$\begin{aligned}P\left(Y_{n}\leq t<Y_{n}+X_{n+1}\right) & =\int_{0}^{\infty}P\left(Y_{n}\leq t<Y_{n}+X_{n+1}\mid Y_{n}=y\right)f_{Y_{n}}\left(y\right)dy\\ & =\int_{0}^{\infty}P\left(y\leq t<y+X_{n+1}\mid Y_{n}=y\right)f_{Y_{n}}\left(y\right)dy\\ & =\int_{0}^{\infty}P\left(y\leq t<y+X_{n+1}\right)f_{Y_{n}}\left(y\right)dy\\ & =\int_{0}^{t}P\left(t<y+X_{n+1}\right)f_{Y_{n}}\left(y\right)dy\\ & =\int_{0}^{t}P\left(X_{n+1}>t-y\right)f_{Y_{n}}\left(y\right)dy \end{aligned} $$

where the third equality rests on independence.

Also be aware that $P\left(y\leq t<y+X_{n+1}\right)=0$ if $y>t$ because in that case: $$\left\{ \omega\in\Omega\mid y\leq t<y+X_{n+1}\left(\omega\right)\right\} =\varnothing$$

Answer 2

$$P(Y_n \leq t <Y_n+X_{n+1})=E[1_\left\{Y_n \leq t <Y_n+X_{n+1} \right\}]=\int_{\mathbb{R}}(\int_{\mathbb{R}}1_\left\{x \leq t <x+y \right\}(x,y)dP_{X_{n+1}}(y))dP_{Y_{n}}(x)=\int_{\mathbb{R}}1_{[0,t]}(x)(\int_{\mathbb{R}}1_{\left\{y>t-x \right\}}(y)dP_{X_{n+1}}(y))dP_{Y_n}(x)=\int_{0}^tP(X_{n+1}>t-x)f_{Y_n}(x)dx$$

using independence and Fubini