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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?

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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

$$ \alpha \ge \beta.$$

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  • $\begingroup$ 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! $\endgroup$ – JCF Mar 26 at 18:57
  • $\begingroup$ 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 $\log \psi$? $\endgroup$ – JCF Mar 27 at 11:48
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    $\begingroup$ divide by $t$, send to $t\to\infty$. $\endgroup$ – Fnacool Mar 27 at 12:06

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