I am solving a question for applied stochastic processes homework and I am stuck on this part:
Let $X_1,X_2,\cdots, X_n$ be independent identically distributed random variables with uniform distribution on $[0,1]$. Let $0<x<1$. I need to find $$\mathbb{P}(X_1+\cdots+X_n<x \mid X_1+\cdots +X_{n-1}<x).$$
I think that the answer should be $\mathbb{P}(X_1+\cdots+X_n<x \mid X_1+\cdots +X_{n-1}<x)=\frac{x}{n}$, for reasons of symmetry , since we have $n$ RV then each of them in average should be less then $\frac{x}{n}$ so that the total sum is less then $x$. Is my guess correct? And how can I state it in more mathematical basis?
First note that
\begin{align*} \Bbb{P}(S_{n} < x) &= \Bbb{P}(S_{n-1} < x - X_{n} \,;\, X_{n} < x) \\ &= \int_{0}^{x} \Bbb{P}(S_{n-1} < x - y) \, dy \\ &= \int_{0}^{x} \Bbb{P}(S_{n-1} < y) \, dy. \end{align*}
Since $\Bbb{P}(S_{1} < x) = x$, by induction we can easily check that
$$ \Bbb{P}(S_{n} < x) = \frac{x^{n}}{n!}. $$
In particular, you get
$$ \Bbb{P}(S_{n} < x \mid S_{n-1} < x) = \frac{x}{n}. $$