Frog Jumps - hard time understanding the probability mass distribution

Question

I got the following riddle:

A frog jumps for $T$ steps starting from $0$, each step being $X_i\sim U (0,1)$, till $$S_T=\sum_{i=1}^T X_i > 1.$$ Compute $\mathbb E[T]$.

The procedure I found everywhere reads:

1. call $\mathbb P(T > t)=\mathbb P(S_t\leq 1)$.

2. we need to compute the probability mass distribution of $T$, call it $\mathbb P(T=t)$, as: \begin{equation} \mathbb P(T=t)=\mathbb P(T > t-1)-\mathbb P(T> t). \label{eq:p}\tag{1}\end{equation}

3. one knows that $\mathbb P(S_t\leq 1)=(t!)^{-1}$ by geometric arguments (volume of a $t$-dimensional simplex).

4. the proof will eventually lead to $\mathbb E[T]=e$.

Would you please explain me why the distribution of $T$ can be written as in Eq.\eqref{eq:p}?

ANSWER:

Imagine, you know that $$\sum_{t=1}^\infty \mathbb P(T=t) = 1,\quad\text{and}\quad\mathbb P(T > t -1) = \mathbb P(T \geq t)=\mathbb P(T > t) + \mathbb P (T=t),$$ so that Eq.\eqref{eq:p} follows.

EDIT:

I would have written $$\tag{2} \label{eq:2}\mathbb P(T=t)=\mathbb P(T>t|S_{t-1}\leq1)\mathbb P(S_{t-1}\leq1),$$ and try (but failing) to compute $\mathbb P(S_t>1|S_{t-1}\leq1)$, but the answer from claimes clarified the mistake.

Answer 1

I think you just have a spin in the equation, as $$\mathbb{P}[T = t] = \mathbb{P}[(T>t-1) \setminus (T>t)] = \mathbb{P}[T>t-1] - \mathbb{P}[T>t].$$ Actually we have not $\mathbb{P}[T \le t] = \mathbb{P}[S_t \le t]$ but $\mathbb{P}[T > t] = \mathbb{P}[S_t \le t]$ which you can see for example by the fact that the probability decreases for increasing t if you would choose the original 1. Therefore we can follow: $$\mathbb{P}[T = t] = \mathbb{P}[S_t \le t-1] - \mathbb{P}[S_t \le t] = \frac{1}{(t-1)!} - \frac{1}{t!} = \frac{t-1}{t!}$$ And now we see $$\mathbb{E}[T] = \sum_{t = 1}^{\infty} \mathbb{P}[T = t]\cdot t = \sum_{t = 1}^{\infty} \frac{t-1}{t!} \cdot t = \sum_{t = 0}^{\infty} \frac{t}{t!} = \sum_{t = 1}^{\infty} \frac{1}{(t-1)!} = \sum_{t = 0}^{\infty} \frac{1}{t!} = e$$

Maybe you understand the given hints now in a better way.

Regarding your equation 2 we got $$\mathbb P(T>t|S_{t-1}\leq1) \cdot \mathbb P(S_{t-1}\leq1) = \mathbb{P}[T > t \, \cap S_{t-1} \le 1] = \mathbb{P}[T > t \, \cap T > t-1] = \mathbb{P}[T > t]$$ which is not the same as $\mathbb{P}[T = t]$