# Decay rate of a birth-death Markov chain and relationship with the support of the orthogonalizing probability

I am studying this Article by van Doorn on the existence of quasi-stationary distribution for a birth-death process with killing.

He defines the decay rate of a birth-death process with killing as $$\alpha =- \lim_{t\rightarrow \infty} \frac{1}{t} \log P_{ij}(t),$$ where $$P_{ij}(t)$$ are the transition probabilities $$P_{ij}(t) = \mathbb{P} (X_t = j | X_0 = i).$$ It can be shown that $$\alpha$$ is independent from $$i,j$$. Moreover there exists also an integral representation for the transition probability under suitable assumption on the process, namely the following one holds $$P_{ij}(t) = K_j \int_0^\infty e^{-xt}Q_{i-1}(x)Q_{j-1}(x)\psi(dx)$$ where $$\{ Q_n \}_{n \in \mathbb{N}}$$ is a orthogonal polynomial sequence given by the parameter of the process, that is not interesting for the question I am going to ask, anyway I underline that $$Q_0 (x) = 1$$. So that if $$i=j=1$$ we have that $$P_{11}(t) = \int_0^\infty e^{-xt} \psi(dx), \quad t \geq 0.$$ Hence the author states that is an easy consequence of the integral representation of $$P_{11}(t)$$ that $$\alpha = \inf \text{supp}( \psi).$$ However I don't get why this would imply $$\alpha$$ to be the infimum of the support of $$\psi$$, any suggestions?

Preparation:

• $$\psi$$ is a probability measure on $$[0,\infty)$$, because $$1=P_{1,1}(0)= \psi([0,\infty))$$.
• Recall that $$x\in [0,\infty)$$ is in the support of $$\psi$$ if and only if $$\psi(U)>0$$ for any neighborhood of $$x$$. Let $$\beta = \inf \mbox{supp}(\psi)$$.
• As a result,
$$(*) \quad\psi\left( [0,\beta)\right)=0.$$

Fix $$x$$ in the support of $$\psi$$. Then

$$P_{1,1}(t) \ge \int_{(x-1/n,b+1/n)\cap[0,\infty)}e^{-xt} \psi(dx) \ge e^{-(x+1/n) t } \psi((x-\frac1n,x+\frac 1n)\cap [0,\infty)).$$

Thus,

$$\liminf \frac{1}{t} \ln P_{1,1}(t) \ge -(x+\frac 1n).$$

Which implies $$\alpha \le x+\frac 1n$$. Since $$x$$ is any point in the support of $$\psi$$ and $$n$$ is arbitrary, it follows that $$\alpha \le \beta$$.

On the other hand by $$(*)$$ we have

$$P_{1,1}(t) = \int_{[\beta,\infty)} e^{-xt} \psi(dx)\le e^{-\beta t} \int_{[\beta,\infty)}\psi(dx) =e^{-\beta t}.$$

Therefore

$$\alpha \ge \beta.$$

• Thank you so much, I surely need to become more fluent with this kind of arguments. I only underline an $x$ that became $b$ in the domain of integration, anyway thank you again! – JCF Mar 26 at 18:57
• applying $\log$ to both members I obtain $\log P_{11}(t) \geq - (x + \frac{1}{n}) t + \log (\psi ( (x- \frac{1}{n}, x+\frac{1}{n}) \cap [0, \infty) )$ how did you eliminate the part with $\log \psi$? – JCF Mar 27 at 11:48
• divide by $t$, send to $t\to\infty$. – Fnacool Mar 27 at 12:06