Extremal distribution of random variable that averages to a given value Denote by $\mathcal M$ the set of probability measures over $[0,1]$ that average to $p$. This is a convex set and I am trying to characterize its extreme points. For any $x$, $y$ such that $0\leq x< p< y\leq 1$ define
\begin{align*}
\lambda_x^y = \frac{p-y}{x-y}\cdot\delta_x+\frac{x-p}{x-y}\cdot\delta_y
\end{align*}
where for any measurable $A\subseteq [0,1]$ and $a\in[0,1]$, $\delta_a(A)=\mathbf 1(a\in A)$. Since $\delta_x$ (resp. $\delta_y$) averages to $x$ (resp. $y$) we get that $\lambda_x^y$ averages to $\frac{p-y}{x-y}\cdot x+\frac{x-p}{x-y}\cdot y=p$ and so is in $\mathcal M$.
I am very tempted to say that the extreme points of $\mathcal M$ is the set $\{ \lambda_x^y : 0\leq x< p < y \leq 1 \}\cup \{ \delta_p \}$ without being able to prove it. If I am not mistaking I am supposed to prove that (ignoring the $\delta_x$ component)

*

*If for some $x$ and $y$ and $a\in[0,1]$, $\lambda_x^y=a \mu+(1-a) \nu$ then $\mu=\nu=\lambda_x^y$.

*For any $\mu\in\mathcal M$ there is a probability measure $\kappa$ over $[0,p]\times[p,1]$ such that $\mu(A)=\int \lambda_x^y(A) d\kappa(x,y)+\mu(\{ p \})\delta_p(A)$ for all measurable $A\subseteq [0,1]$.

For the first point, it is quite clear by positivity of the measures $\mu$ and $\nu$ that their support $\{ x,y \}$, then there is only one probability measure with such a support that averages to $p$ and it is $\mu_x^y$.
For the second statement however it is less clear to me, Maybe one way to proceed is computing the integral
\begin{align*}
\int \lambda_x^y(A) d\kappa(x,y)&=\int \left( \frac{p-y}{x-y}\cdot\delta_x(A)+\frac{x-p}{x-y}\cdot\delta_y(A) \right) d\kappa(x,y)\\
&=\int_{\left(A\cap [0,p]\right)\times [p,1]}\frac{p-y}{x-y} d\kappa(x,y)+\int_{[0,p]\times\left(A\cap [p,1]\right)}\frac{x-p}{x-y}d\kappa(x,y)
\end{align*}
It now feel easier to separate cases $A\subseteq [0,p]$ and $A\subseteq [p,1]$ to set one of the two integral to $0$ but I still don't know how to proceed.
Actually thinking about how I could use it in my research it would be much more powerful to not enforce $x\leq p \leq y$ but just that $p$ is in the convex hull of $\{ x,y \}$ and then we get repetition $\lambda_x^y=\lambda_y^x$ but the of extreme points is the same. This makes $\kappa$ a probability measure over $[0,1]^2$ where we have to be careful assigning $0$ probability to sets $A\times B\subseteq [0,1]^2$ where the convex hull of $A\cup B$ does not contain $p$.

Here is the last thing I tried :
Denote $a_x^y=\frac{p-y}{x-y}$, for $\mu \in\mathcal M$ and $A$ measurable
\begin{align*}
\mu(A)&=\int_{[0,p)} \delta_x(A) d\mu(x)+\int_{(p,1]} \delta_y(A) d\mu(y)+\mu(\{ p \})\delta_p(A)\\
&=\int_{[0,p)\times (p,1]} \left(\frac{a_x^y}{a_x^y}\delta_x(A) + \frac{1-a_x^y}{1-a_x^y}\delta_y(A) \right)d\mu\otimes\mu(x,y)+\mu(\{ p \})\delta_p(A)\\
\end{align*}
It feels like we can define $\kappa$ as a function of $\mu\otimes \mu$ and $a_x^y$, I cannot finish the argument though.

Your identificaton of the extreme points of $\mathcal M$ is correct.  A proof of the difficult part can be based on a theorem of R.G. Douglas. See my recent answer to Two term martingales and their extreme points, with a reference to a paper of A.F. Karr containing such a proof.
