I'm trying to find the expectation of a stopping time. Specifically,
Let $T_1,...,T_n$ be i.i.d exponential random variables with mean $1$. Let $S_n = T_1 + ... + T_n$ denote their partial sum. Define the stopping time,
$$ T = \inf\{n \geq 1 : S_n \geq 1\}$$
which is the first time $S_n$ exceeds $1$. Calculate $E[T]$.
Here was my approach. Let $x = E[T]$. I want to condition on what happens at the first time $T_1$. If $T_1 \geq 1$, then the process, $\{S_n\}$ has stopped and $T \equiv 1$. Otherwise if $T_1 < 1$, then after one step, process, $\{S_n\}$ will renew again until it reaches $1$. So we have the following equation,
\begin{eqnarray} x &=& E[T]\\ &=& E[T | T_1 < 1]P(T_1 < 1) + E[T | T_1 \geq 1]P(T_1 \geq 1)\\ &=& (1+x)P(T_1 < 1) + P(T_1 \geq 1)\\ &=& (1+x)(1-e^{-1}) + e^{-1} \end{eqnarray}
This results in $x = e$.
However, I was told that the answer is $2$. They gave a heuristic explanation that $S_T$ is distributed as $1 + S$ where $S$ is an exponential random variable with mean $1$. I can see the intuition behind this from the memoryless property, but I can't prove why it is so. I ran three simulations in $\textsf{R}$ and got $x \approx 2.001, 2.0161, 1.9785$, which seems to confirm that the answer is $2$. Can someone explain this result?
Also, why/where did my approach fail?
Following @stochasticboy321's approach, we want to find $P(T > n) = P(S_n < 1)$. Since $S_n = T_1 + ... + T_n \sim$ Gamma($n,1$), we have,
$$ P(S_n < 1) = \frac{1}{\Gamma(n)}\int_0^1x^{n-1}e^{-x}dx = 1 - \frac{\Gamma(n,1)}{\Gamma(n)}$$
where $\Gamma(n,1)$ is the incomplete Gamma function. To get this expression, I used the nice identity found here. Finally,
$$ E[T] = 1 + \sum\limits_{n=1}^\infty P(T > n) = 1 + \sum\limits_{n=1}^\infty P(S_n<1) = 1+ \sum\limits_{n=1}^\infty\left(1 - \frac{\Gamma(n,1)}{\Gamma(n)}\right)$$
I have no idea how to evaluate the sum in closed form, but a computation in Wolfram Alpha seems to suggest it converges to $1$. Thus, $E[T] = 2$ (at least by conjecture from this numerical computation).