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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.

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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.

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  • $\begingroup$ Iosif, thanks a lot. What's missing is to show that the entries of the Jacobian matrix are all non-negative. Or is this trivial somehow? $\endgroup$
    – lum
    Jun 10 '15 at 8:33
  • $\begingroup$ As for your comments: 1) + 2) True. In my case the limit exists and is differentiable. 3) You are right, I changed my definition of the Jacobian matrix accordingly. Also, I guess you mean $\mathbf{p}Q_{\mathbf{p}} = [t_{1;\mathbf{p}}, \ldots, t_{n;\mathbf{p}}]$. $\endgroup$
    – lum
    Jun 10 '15 at 8:36
  • $\begingroup$ In my answer, I have fixed the typo concerning $\mathbf{p}Q_\mathbf{p}$ and added a comment about the (lack of) nonnegativity, in general. $\endgroup$ Jun 10 '15 at 15:52
  • $\begingroup$ But the Jacobian matrix of $\mathbf{p}P$ is row-stochastic by assumption. Can this help? $\endgroup$
    – lum
    Jun 10 '15 at 17:05
  • $\begingroup$ Reading your question, I had thought that you proved that the Jacobian matrix of pP is row-stochastic, instead of assuming it is so. Anyway, even this assumption does not help, as shown in the added, last paragraph of my answer. $\endgroup$ Jun 10 '15 at 20:12

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