Equivalence of two methods for generating random numbers that sum to 1 I want a vector $\vec{v}$ of $N$ non-negative random numbers that sum to 1.
Let $X(a)$ be the (continuous) uniform distribution over interval $[0, a]$.
Let $S(n) = \sum_{i = 1}^{n} v_{i}$ be the partial sum of the elements of $\vec{v}$
Method 1

*

*Generate: For each $k$, set $v_k$ to a random number from $X(1)$.

*Normalize: Divide $\vec{v}$ by sum of all elements of $\vec{v}$.

Method 2
Generate the elements of $\vec{v}$ one after another with the following steps.

*

*Generate 1st element: set $v_1$ to a random number from $X(1)$.


*Generate 2nd element: set $v_2$ to a random number from $X(1 - v_{1})$
...


*Generate the $k^{th}$ element: set $v_k$ to a random number from $X(1 - S(k-1))$.
...


*Calculate the last element: set $v_N$ to $1 - S(N - 1)$.
Question
Are the two methods equivalent?
Do the two methods generate $\vec{v}$ with the same $N$ dimensional probability density?
Thank you.
 A: These methods are certainly not equivalent. It is easy to see that in the first method, $\mathbb{E}[v_n]=\frac 1n$, while in the second method, $\mathbb{E}[v_k]=\frac 1{2^k}$ (you can prove this with induction and linearity of expectation), and so $\mathbb{E}[v_n]=\frac 1{2^n}$.

In response to a comment on another answer, scrambling the elements of the array also does not make the methods equivalent.
With method 2, there is a $\frac 13$ probability that there exists an element with value greater than $\frac 23$ (this is the element that was chosen first).
With method 1, there is only a small probability that some element has value greater than $\frac 23$. To show this, let the vector before normalization be $u$, so that $v=\frac u{\sum u}$. Let us consider the case where element $n$ has value greater than $\frac 23$. Clearly this happens only if $\sum_{i=1}^{n-1}u_i \leq \frac {u_n}2 \leq \frac 12$. But if $\sum_{i=1}^{n-1}u_i \leq \frac 12$, then $u_i \leq \frac 12 \forall i \in [1\dots n-1]$, and clearly this only happens with probability $\frac 1{2^{n-1}} \leq \frac 13$. (Union bounding only changes this to $\frac n{2^{n-1}}$, which is $\leq \frac 13.$)
A: The answer to this is that these do not give equivalent distributions if the order of your vector matters (i.e. the entries are considered in-exchangeable).
To see this, we note that under Method 1, any of the $N$ samples is equally likely to be the biggest sample. In particular this means for $2 \leq n \leq N$:
$$ \mathbf P[ v_1 \geq v_n] = \frac12.$$
Now considering Method 2, there is a natural order to the problem that implies that on average earlier samples $v_k$ will be larger than later ones.
Consider the case $N = 3$, and note that
$$v_3 = 1 - v_1 - v_2$$
Then 
$$
\begin{aligned}
\mathbf P[v_1 \geq v_3] &= \mathbf P[v_2 \geq 1-2v_1] \\
& =\int_{0}^1 \mathbf P[v_2 \geq 1 - 2v_1 | v_1 = s] \mathbf P[v_1 = s] ds \\
& = \int_{0}^1 \mathbf P[v_2 \geq 1 - 2s |v_1=s] ds \\
& = \int_{0}^1 \left\{ 1 - \mathbf P[v_2 \leq 1 - 2s |v_1=s]\right\} ds \\
& = \int_{0}^1 \left\{1 - P(v_2 \leq 1 - 2s|v_2 \sim\text{Unif}[0,1-s]) \right\}ds\\
& = 1 - \int_0^1 \frac{1-2s}{1-s}\mathbf 1\left(0 \leq 1-2s \leq 1-s\right)ds
\\
& = 1 - \int_0^{1/2} \frac{1-2s}{1-s}ds \\
& = \log(2) \\
\end{aligned}
$$
In particular we have $P[v_1 \geq v_3] > \frac12$, so the two distributions do not agree.
