How to 'randomize' a given discrete probability distibution? Given a discrete probability distribution, $\boldsymbol{\lambda} = (\lambda_1,\lambda_2, ..., \lambda_n)$, how can we construct a random distribution,
$\boldsymbol{\Lambda} = (\Lambda_1, \Lambda_2, ..., \Lambda_n)$ such that $\mathbb{E}(\boldsymbol{\Lambda}) = \boldsymbol{\lambda}$?
The goal is to generate random variations of the initial distribution, $\boldsymbol{\lambda}$, with the above property. If it is possible to do this and have some degree of control over the variance/spread of each $\lambda_i$, then even better!
I suspect that there are many ways to achieve this, so any approaches/answers are appreciated.
What I've tried:
My initial thought was to let each $\Lambda_i \sim Beta(\alpha_i, \beta_i)$, with $\alpha_i$, $\beta_i$ chosen so that $\mathbb{E}(\Lambda_i) = \lambda_i$. By choosing larger $\alpha_i$, and $\beta_i$ values, we can also make the distributions tighter around the mean.
However, this has the issue that, in general,
$$
\sum_{i=1}^n \Lambda_i \ne 1
$$
We may work around this by normalizing $\boldsymbol{\Lambda}$ after evaluating each $\Lambda_i$ (by dividing through by sum). Unfortunately I think I am right in saying that, in general, this means that $\mathbb{E}(\boldsymbol{\Lambda}) = \boldsymbol{\lambda}$ no longer holds.
Similarly, we could 'normalize' $\boldsymbol{\Lambda}$ by setting
$$
\Lambda_n = 1 - \sum_{i=1}^{n-1} \Lambda_i
$$
This preserves $\mathbb{E}(\boldsymbol{\Lambda}) = \boldsymbol{\lambda}$ (at the expense of the distribution of $\Lambda_n$ now being difficult to determine exactly, though we can approximate it). The problem with this method is that $\Lambda_n$ could now be negative!
 A: Rather embarrassingly, I seem to have overlooked the Dirichlet distribution!
Nevertheless, any other approaches, would be much appreciated! In particular, methods of generating $\boldsymbol{\Lambda}$ that give easier control over the spread/variance of each $\Lambda_i$ would be useful.
A: You could generate a random vector that sums to $1$ and subtract $\frac1n$ from each component, then add some multiple of the resulting zero-sum vector to $\boldsymbol{\lambda}$. A straightforward way to generate a random vector that sums to $1$ while keeping all components on an equal footing is to use the $n$ intervals formed by $n-1$ points independently uniformly chosen in the unit interval. Then $\mathbb{E}(\boldsymbol{\Lambda}) = \boldsymbol{\lambda}$ follows by symmetry. You could ensure that the $\Lambda_i$ lie in $[0,1]$ by adding a sufficiently small multiple of the random vector. (The factor would need to be determined beforehand, independent of the random vector, in order to preserve $\mathbb{E}(\boldsymbol{\Lambda}) = \boldsymbol{\lambda}$.)
