Define $F(z) : \mathbb{R} \to [0,1]^2$, $C \in \mathbb{R}^{2 \times 2}$ and $D \in \mathbb{R}^{2}$. The scalar $z \in [0, \infty)$
A few further assumptions:
- $C$ is negative definite and diagonalizable, but not symmetric
- $\min(eig(-C)) > 1$, but that is likely overkill here.
- $D > 0$
- $F(z)$ has full support on $[0,\infty)$
- If it is useful, the sum of $-C^{-1} D$ is 1
Short Version of the Question:
Define the PDF using the matrix exponential. $$F'(z) = e^{C z} D$$
The question is: under what additional conditions on $C$ and $D$ is this a valid PDF? In particular,
- $\lim_{z\to\infty}F'(z) = 0$
- $F'(z) > 0$ for all $z$.
I believe the 1st condition only requires a negative definite $C$. But see an example at the bottom of this question shows that this is not enough for the 2nd condition
An approach?
As the limit must go to $0$ and it must be positive, is it sufficient to prove monotonicity of $F'(z)$? This is certainly true with any reasonable solutions to the problem. If so, can we simply prove for all $z \geq 0$ that $$ F''(z) < 0 $$ From the definition, doesn't this mean proving for all $z$ that the following has no solution for $z\geq 0$? $$ e^{C z} C D = 0 $$ If we do the diagonalization, $C \equiv Q B Q^{-1}$ then this becomes, $$ Q e^{B z} Q^{-1} C D = 0 $$ But I couldn't figure out anything more after this...(This could be related to Can matrix exponentials ever be negative? If so, under what conditions?)
Background: I have a two-state Markov chain and a continuous state, and have a Kolmogorov forward equation to describe the stationary CDF. then the ODE can be written as: $$F'(z) = C F(z) + D$$ Where $z \in [0,\infty)$ and initial condition $F(0) = 0$ (and $C$ and $D$ will fulfill some conservation of mass $F_1(\infty) + F_2(\infty) = 1$)
The solution to this ODE for the CDF is: $$ F(z) = (e^{C z} - \mathbb{I})C^{-1}D $$ With the PDF: $$ F'(z) = e^{C z} D $$
From this, you can see why $\sum -C^{-1} D = 1$ to ensure that $\sum \lim_{z\to\infty} F(z) = 1$
Example (one I want to reject) Writing in Matlab to simplify the copy/paste, define the following example
C = [- 2.2959 -1.5; -.1 -1.6918];
min(eig(-C)) %Can check the eigenvalues to ensure negative definite
D = [2.0; 0.6918];
F_p = @(z) expm(C * z) * D;
%Evaluate at a few z's
F_p(1) %Both > 0
F_p(3) %One of them < 0!!!
F_p(100) %They both converge to 0
%Could diagonalize C
[Q,B] = eig(C)
% Then define F''(z)
F_pp = @(z) Q * diag(exp(diag(B) * z))* inv(Q) * C * D