Find the distribution of $T_a=\inf\{n\ge 0: R_{n}\gt a\}$ for fixed number $a\gt 0$

Question

Suppose $R_{n}=\sum_{i=1}^{n} X_{i}$ for $n\ge 1$ and $R_{0}=0$ , where $X_{i}\gt 0$ are independent and identically distributed. Find the probability law of the stopping time $T_a=\inf\{n\ge 0: R_{n}\gt a\}$ for fixed number $a\gt 0$.

Answer 1

Well I'm not Did sorry about that.

What you want to do is getting $\forall k>0$, $P(T_a\le k)$, but $\{T_a\le k\}$ is the same event as $\{R_k>a\}$ (as $R_k$ is an increasing function in $k$). So $P(T_a\le k)=P(R_k >a)$.

Now the law of the sum of independent random variables is given for $k>3$ by the following recursive integration formula (I hope I haven't made mistakes here):

$P(R_k>a)=\int_0^a(...(\int_0^{x_{k-3}}P[x_{k-1}+x_{k}>x_{k-2}].dP_X(x_{k-2}))...)dP_X(x_1)$

with $P[x_{k-1}+x_{k}>x]=\int_{\mathbb{R^+}^2}1[x_{k-1}+x_{k}>x]dP_X(x_{k-1}).dP_X(x_{k})$

and everywhere $P_X$ is $X_i$'s probability law.

To be exhaustive here are the cases for other values of $k$.

if $k=3$ then : $P(R_k>a)=\int_0^aP[x_{2}+x_{3}>x_{1}].dP_X(x_{1})$

if $k=2$ : $P[x_{1}+x_{2}>a]=\int_{\mathbb{R^+}^2}1[x_{1}+x_{2}>a]dP_X(x_{1}).dP_X(x_{2})$

if $k=1$ : $P_X(x_{1}>a)$

Best regards