What is this matrix of binomial distributions? I have come across the following type of $(n+1) \times (n+1)$ matrix:
$$C^n = \left( \begin{array}{cccc}
 \beta^n(0) & 0   & \dots & 0  \\
 \beta^n(1) & \beta^{n-1}(0) & \dots & 0 \\
 \vdots & \vdots  & \ddots & \vdots \\
 \beta^n(n) & \beta^{n-1}(n-1)& \dots & \beta^0(0) \\
 \end{array} \right)$$
where $\beta^n(k) = {n \choose k} p^k(1-p)^{n-k}, k = 0,1,\dots, n$ is the vector of binomial distribution probabilities, and we define $\beta^0(0) = 1$. For example, for $n=3$ we have the matrix
$$ C^3 = \left( \begin{array}{cccc}
 (1-p)^3  & 0   & 0   & 0 \\
 3 p(1-p)^2 & (1-p)^2  & 0  &0 \\
 3 p^2(1-p) & 2 p(1-p) & 1-p  & 0 \\
 p^3   & p^2  & p  & 1 \\
 \end{array} \right) $$
This matrix seems to have some interesting properties: it is clearly invertible; the lower right $n\times n$ submatrix of $C^n$ is $C^{n-1}$; and it appears that these matrices commute with convolution of probabilities, in the sense that
$$  C^{m+n} (x \otimes y) = (C^m x) \otimes (C^n y)$$
where $x \in \mathbb{R}^n$ and $y\in \mathbb{R}^m$ are discrete probability distribution vectors and $x \otimes y$ is the usual convolution operation.
Is anyone familiar with this kind of matrix? I would be grateful for any suggestions for literature/information. Have been searching for most of the day, but I'm probably not using the right keywords ... I am aware of the related Pascal matrices, but this seems a bit different.
 A: Let's invert the rows/columns of your matrix and adopt the following notation
$$
{\bf P}_{\,h} (p) = \left\| {\,P_{\,n,\,m} (p)\;\left| {\;0 \le n,m \le h} \right.\;} \right\|\;:\;P_{\,n,\,m} (p)
   = \binom{n}{m}p^{\,n - m} \left( {1 - p} \right)^{\,m} 
$$
which is
$$
{\bf P}_{\,h} (p) = \left\| {\matrix{
   {\left( \matrix{
  0 \cr 
  0 \cr}  \right)p^{\,0} \left( {1 - p} \right)^{\,0}  = 1} & 0 & 0 &  \cdots  & 0  \cr 
   p & {\left( {1 - p} \right)} & 0 &  \cdots  & 0  \cr 
   {\left( \matrix{
  2 \cr 
  0 \cr}  \right)p^{\,2} } & {\left( \matrix{
  2 \cr 
  1 \cr}  \right)p\left( {1 - p} \right)} & {\left( \matrix{
  2 \cr 
  2 \cr}  \right)\left( {1 - p} \right)^{\,2} } &  \cdots  & 0  \cr 
    \vdots  &  \vdots  &  \vdots  &  \ddots  &  \vdots   \cr 
   {\left( \matrix{
  h \cr 
  0 \cr}  \right)p^{\,h} } & {\left( \matrix{
  h \cr 
  1 \cr}  \right)p^{\,h - 1} \left( {1 - p} \right)} & {\left( \matrix{
  h \cr 
  2 \cr}  \right)p^{\,h - 2} \left( {1 - p} \right)^{\,2} } &  \cdots  & {\left( \matrix{
  h \cr 
  h \cr}  \right)\left( {1 - p} \right)^{\,h} }  \cr 
 } } \right\|
$$
We make this change in order to work with the "standard" Pascal Matrix 
$$
{\bf B}_{\,h}  = \left\| {\;B_{\,n,\,m}  = \binom{n}{m}\;\left| {\;0 \le n,m \le h} \right.\;} \right\|
$$
since we can write
$$
\eqalign{
  & {\bf P}_{\,h} (p) = \left\| {\,\binom{n}{m}p^{\,n - m} \left( {1 - p} \right)^{\,m} \;} \right\|\;
 = \left( {p^{\,n}  \circ {\bf I}_{\,h} } \right){\bf B}_{\,h} \left( {\left( {{{1 - p} \over p}} \right)^{\,n}  \circ {\bf I}_{\,h} } \right)  \cr 
 &  = \left( {p^{\,n}  \circ {\bf I}_{\,h} } \right){\bf B}_{\,h} \left( {q^{\,n}  \circ {\bf I}_{\,h} } \right)\left( {p^{\,n}  \circ {\bf I}_{\,h} } \right)^{\, - \,{\bf 1}}
   = \left( {p^{\,n}  \circ {\bf I}_{\,h} } \right){\bf B}_{\,h} \left( {p^{\,n}  \circ {\bf I}_{\,h} } \right)^{\, - \,{\bf 1}} \left( {q^{\,n}  \circ {\bf I}_{\,h} } \right) =   \cr 
  &  = {\bf B}_{\,h} ^{\,{\bf p}} \left( {q^{\,n}  \circ {\bf I}_{\,h} } \right) = \left( {q^{\,n}  \circ {\bf I}_{\,h} } \right){\bf B}_{\,h} ^{\,{\bf p}/{\bf q}}  =  \cdots  \cr} 
$$
where by
$$
\left( {f(n) \circ {\bf I}_{\,h} } \right)
$$
I indicate the diagonal matrix having the diagonal non null elements equal to $f(n)$.
Your matrix will then correspond to
$$
\eqalign{
  & {\bf C}_{\,h} (p) = \left\| {\;C_{\,n,\,m} (p) = \binom{h-m}{h-n}p^{\,n - m} \left( {1 - p} \right)^{\,h - n} \;} \right\| =   \cr 
  &  = {\bf J}_{\,h} \left( {\left( {1 - p} \right)^{\,n}  \circ {\bf I}_{\,h} } \right){\bf J}_{\,h} \left( {p^{\,n}  \circ {\bf I}_{\,h} } \right){\bf J}_{\,h} \,\overline {{\bf B}_{\,h} }
 \,{\bf J}_{\,h} \left( {p^{\,n}  \circ {\bf I}_{\,h} } \right)^{\, - \,{\bf 1}}   \cr 
  &  = {\bf J}_{\,h} \,\overline {{\bf P}_{\,h} (p)} \,{\bf J}_{\,h}  \cr} 
$$
where the overbar indicates the transpose, and $\bf J$ the exchange matrix.
There are many properties that can be derived concerning the matrices $\bf P , \; \bf C$.
The first is that both are stochastic, as already noted
$$
{\bf u}_{\,h}  = \left\| {\,\matrix{   1  \cr    1  \cr     \vdots   \cr    1  \cr  } \,} \right\|
 = {\bf P}_{\,h} (p)\;{\bf u}_{\,h} \quad \overline {{\bf u}_{\,h} }  = \overline {{\bf u}_{\,h} } \;{\bf C}_{\,h} (p)
$$
The second is that, indicating as usual $q=1-p$, we clearly have
$$
charpoly\left( {{\bf P}_{\,h} (p)} \right) = charpoly\left( {{\bf C}_{\,h} (p)} \right) = \prod\limits_{k = 0}^h {\left( {x - q^{\,k} } \right)} 
$$
and the eigenvalues and the determinant are easily deducible.
Then since it is
$$
\eqalign{
  & {\bf P}_{\,h} (p)\,{\bf B}_{\,h}  = \left( {q^{\,n}  \circ {\bf I}_{\,h} } \right){\bf B}_{\,h} ^{\,{\bf p}/{\bf q}} \,{\bf B}_{\,h}  =   \cr 
  &  = \left( {q^{\,n}  \circ {\bf I}_{\,h} } \right){\bf B}_{\,h} ^{\,\left( {{\bf p} + {\bf q}} \right)/{\bf q}}  = \left( {q^{\,n}  \circ {\bf I}_{\,h} } \right){\bf B}_{\,h} ^{\,{\bf 1}/{\bf q}}  =   \cr 
  &  = \left( {q^{\,n}  \circ {\bf I}_{\,h} } \right)\left( {\left( {{1 \over q}} \right)^{\,n}  \circ {\bf I}_{\,h} } \right){\bf B}_{\,h} 
\left( {\left( {{1 \over q}} \right)^{\,n}  \circ {\bf I}_{\,h} } \right)^{\, - \,{\bf 1}}  =   \cr 
  &  = {\bf B}_{\,h} \left( {q^{\,n}  \circ {\bf I}_{\,h} } \right) \cr} 
$$
it comes out that the eigenvectors of $\bf P$ are given by the Pascal matrix $\bf B$
and that $\bf P$ diagonalize as
$$ \bbox[lightyellow] {  
\eqalign{
  & {\bf P}_{\,h} (p)\, = {\bf B}_{\,h} \left( {q^{\,n}  \circ {\bf I}_{\,h} } \right){\bf B}_{\,h} ^{\, - \,{\bf 1}}   \cr 
  & {\bf C}_{\,h} (p) = {\bf J}_{\,h} \,\overline {{\bf P}_{\,h} (p)} \,{\bf J}_{\,h}
  = \left( {{\bf J}_{\,h} \overline {{\bf B}_{\,h} } ^{\, - \,{\bf 1}} } \right)\,\left( {q^{\,n}  \circ {\bf I}_{\,h} } \right)\,\left( {{\bf J}_{\,h} \overline {{\bf B}_{\,h} } ^{\, - \,{\bf 1}} } \right)^{\, - \,{\bf 1}}  \cr} 
 }$$
From here it is easy to deduce the powers of $\bf P ,\; \bf C$, as well the functions developable into series.   
Therefrom also follows the interesting, and expected, property:
$$
{\bf P}_{\,h} (1 - a)\;{\bf P}_{\,h} (1 - b) = {\bf P}_{\,h} \left( {1 - ab} \right)
$$
and same for $\bf C$, 
which suggests that we have better exchange $p$ and $q$ in the definition of $\bf P$.
Finally, looking at $\bf P$ as a transition matrix, it is apparent that if we define the single-step matrix $\bf Q$ as
$$
{\bf Q}_{\,h} (p) = \left\| {\;\matrix{
   1 & 0 & 0 &  \ldots   \cr 
   p & {1 - p} & 0 &  \ddots   \cr 
   0 & p & {1 - p} &  \ddots   \cr 
    \vdots  &  \ddots  &  \ddots  &  \ddots   \cr 
 } \;} \right\|
$$
whose meaning is obvious , then the $k$-th row of $\bf P$ is the $k-th$ row of ${\bf Q}^{\,{\bf k}}$, i.e.
$$
{\bf P}_{\,h} (p) = \sum\limits_{0\, \le \,k\, \le \,h} {\left( {\left[ {n = k} \right] \circ {\bf I}_{\,h} } \right){\bf Q}_{\,h} ^{\,{\bf k}} (p)} 
$$
where $[P]$ denotes the Iverson bracket, and
which is the same as saying that
$$
{1 \over {\left( {h + 1} \right)}}\left\| {\;1,1, \cdots ,1\;} \right\|\;{\bf P}_{\,h} (p) = {1 \over {\left( {h + 1} \right)}}\sum\limits_{0\, \le \,k\, \le \,h} {\left\| {\;\left[ {0 = k} \right],\left[ {1 = k} \right], \cdots ,\left[ {h = k} \right]\;} \right\|{\bf Q}_{\,h} ^{\,{\bf k}} (p)} 
$$
the meaning of which is also obvious.
