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


*

*The limit $Q_{\mathbf p}:=\lim_m P_{\mathbf p}^m$ may not exist in general.  

*It is unclear if the limit is differentiable in $\mathbf p$, even when it exists and $P_{\mathbf p}$ is differentiable in $\mathbf p$. 

*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. 
