Let's presume that we have an experiment:
We have $sum = 0$ and we draw value from distribution $U[0, 1]$ and we add it to $sum$ and we continue to do so until $sum>2$.
Now assuming that $\sum_{i=1}^n X_i \le 2$ and for $\sum_{i=1}^{n+1} X_i > 2$:
- how to compute expected $n$ or even better to get distribution for $n$
- how to compute $\mathbb E[X_{n+1}]$?
Let $U_n$ be a sequence of independent variables with uniform distribution over $(0,1)$, and set $X_n = \sum_{i=1}^n X_i$. Let $\tau = \inf\{n>0: X_n>2\}$. To compute the distribution of $\tau$, we must first consider the distribution of $X_n$. For $n=2$ we can compute the density by convolution: \begin{align} f_U\star f_U(t) &= \int_{\mathbb R}f_U(\tau)f_U(t-\tau)\ \mathsf d\tau\\ &= \int_0^t \ \mathsf d\tau \mathsf 1_{(0,1)}(t) + \int_{2-t}^t \ \mathsf d\tau \mathsf 1_{(1,2)}(t)\\ &= t\ \mathsf 1_{(0,1)}(t) + (2-t)\ \mathsf 1_{(1,2)}(t). \end{align} In general $X_n$ has the Irwin-Hall distribution, with density $$ f_n(t) = \frac1{2(n-1)!}\sum_{k=0}^n \binom nk (-1)^k(t-k)^{n-1}\mathrm{sgn}(t-k), $$ where $\mathrm{sgn}(\cdot)$ denotes the sign function. For $n=3$ we see that $$ f_3(t) = \frac12 t^2\mathsf 1_{(0,1)}(t) + \frac12 (-2t^2+6t-3)\mathsf 1_{(1,2)}(t) + \frac12 (t-3)^2\mathsf 1_{(2,3)}(t), $$ and so $$ \mathbb P(\tau = 3) = \mathbb P(X_3>2) = \int_2^3 f_3(t)\ \mathsf dt = \frac16. $$ The distribution function of $X_n$ is given by $$ F_n(t) = \frac1{n!} \sum_{k=0}^{\lfloor t\rfloor}\binom nk(-1)^k(t-k)^n, $$ and so for $n\geqslant 3$, \begin{align} \mathbb P(\tau=n) &= \mathbb P(X_n>2)-\mathbb P(X_{n-1}>2)\\ &= 1-F_n(2)-(1-F_{n-1}(2)) = 2-(F_n(2)+F_{n-1}(2))\\ &= \frac{\left(2^{n-1}-n\right) (n-2)}{n!}. \end{align} The expected value of $\tau$ is \begin{align} \mathbb E[\tau] &= \sum_{n=3}^\infty n\ \mathbb P(\tau=n)\\ &= \sum_{n=3}^\infty \frac{\left(2^{n-1}-n\right) (n-2)}{(n-1)!}\\ &= e(e-1). \end{align}