Let $P\_{\mathbf{p}}$ be a $n \times n$ row-stochastic matrix whose entries are a function of a probability vector $\mathbf{p} \in \mathbf{R}\_{> 0}^n$, $\sum\_i p\_i = 1$ and define the following mapping:
$$
T(\mathbf{p}) = \mathbf{p} \lim\_{m \to \infty} P\_{\mathbf{p}}^m.
$$
Note that $T(\mathbf{p})$ can be seen as the stationary distribution of the Markov chain with transition matrix $P\_\mathbf{p}$. I would like to show that the Jacobian matrix of this mapping, that I define as
$$
(\*) \;\; T'\_{ij} = \frac{\partial T\_j}{\partial p\_i},
$$
is also row-stochastic. So far, I managed to show that the Jacobian matrix of the (simpler) mapping:
$$
\tilde{T}(\mathbf{p}) = \mathbf{p} \ P\_{\mathbf{p}}
$$
is row-stochastic.


Question: is there some way I can use this knowledge in my (bigger) problem above? Maybe some clever application of a chain rule for derivatives?


One approach I've been thinking about is an induction on $m$, but it seems quite tedious.


### Edit on 06.10


Slightly changed my definition of Jacobian matrix $(\*)$, using the transpose now.



Row-stochasticity of the Jacobian matrix of a stationary distribution
There are a few issues here: 


1. The limit $Q\_{\mathbf p}:=\lim\_m P\_{\mathbf p}^m$ may not exist in general.
2. It is unclear if the limit is differentiable in $\mathbf p$, even when it exists and $P\_{\mathbf p}$ is differentiable in $\mathbf p$.
3. It is unclear how you define the Jacobian (matrix?). For it be row-stochastic, in your definition of the Jacobian matrix $i$ has to index the columns, and $j$ has to index the rows.


Once all these issues are cleared, that is, if the limit $Q\_{\mathbf p}$ exists and is differentiable in $\mathbf p$, then one can deal with $\mathbf{p}Q\_{\mathbf p}$ in exactly the same way as one deals with $\mathbf{p}P\_{\mathbf p}$. Namely, writing $\mathbf p=[p\_1,\dots,p\_n]$, $Q\_{\mathbf p}=(q\_{i,j;\mathbf p})\_{i,j=1}^n$, and $\mathbf{p}Q\_{\mathbf p}=[t\_{1;\mathbf p},\dots,t\_{n;\mathbf p}]$, one has 
$$t\_{j;\mathbf p}=\sum\_i p\_i q\_{i,j;\mathbf p}
$$
and hence
$$\sum\_j\frac{\partial t\_{j;\mathbf p}}{\partial p\_k}
=\sum\_j q\_{k,j;\mathbf p}
+\sum\_j\sum\_i p\_i\frac{\partial q\_{i,j;\mathbf p}}{\partial p\_k}
=1
+\sum\_i p\_i\frac{\partial}{\partial p\_k}\sum\_j q\_{i,j;\mathbf p}=1,
$$
since $\sum\_j q\_{i,j;\mathbf p}=1$. 


However, in general one cannot guarantee that the elements of the Jacobian matrix of $\mathbf{p}Q\_{\mathbf p}$ be all nonnegative, even if it is assumed that all the elements of the Jacobian matrix of $\mathbf{p}P\_{\mathbf p}$ are positive. E.g., suppose that $\mathbf{p}=[s\ \;t]$ and $P=\begin{bmatrix} \frac{15}{16}-\frac{s}{4} & \frac{s}{4}+\frac{1}{16} \\ \frac{1}{8} & \frac{7}{8}\end{bmatrix}$ (I am dropping the subscript ${}\_{\mathbf p}$). Then the stationary matrix $Q=\begin{bmatrix} \frac{2}{4 s+3} & \frac{4 s+1}{4 s+3} \\ \frac{2}{4 s+3} & \frac{4 s+1}{4 s+3}\end{bmatrix}$ and $\dfrac{\partial t\_1}{\partial s}=\dfrac{6-8 t}{(4 s+3)^2}<0$ if $\frac34<t<1$, whereas all the elements of the Jacobian matrix $\begin{bmatrix}\frac{1}{16} (15-8 s) & \frac{1}{8} \\
 \frac{1}{16} (8 s+1) & \frac{7}{8}\end{bmatrix}$ of $\mathbf{p}P\_{\mathbf p}$ are positive. 


