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For a Galton-Watson process, replacing an individual's reproduction by a one-descendant uneventful step with probability $q<1$ changes its offspring probability generating function to the displayed mixture. Its fixed-point equation reduces to $G(z)=z$, so the smallest fixed point and hence the extinction probability of a branching process are unchanged. The mean is $q+(1-q)G^{\prime }(1)$, which also preserves the subcritical/critical/supercritical classification. The excluded case $q=1$ never reproduces or dies and need not share the extinction probability.

An uneventful day leaves one aphid, a one-offspring day leaves the parent plus its offspring, and a two-offspring day leaves three aphids. Death leaves none. Thus the daily descendant counts are $0,1,2,3$ with probabilities $(1-q)t$, $q$, $(1-q)r$, $(1-q)s$, respectively. The mean and probability generating function areIndependence across individuals and days makes the daily population a Galton-Watson process, with each surviving parent counted as one of the next day's descendants.

The standard Galton-Watson extinction fixed point result states that the eventual extinction probability of a branching process $e$ from one ancestor is the smallest root in $[0,1]$ of $G(e)=e$. Moreover a nondegenerate offspring law of mean at most one has extinction probability of a branching process one. Here, since $q<1$, the fixed-point equation reduces towhich contains no $q$. This proves laziness invariance of branching extinction: uneventful days change the timescale, not eventual extinction. The excluded case $q=1$ would keep one aphid forever.

If $2r+3s\le 1$, the offspring mean is at most one. In this parameter range $t>0$: if $t=0$, then $r+s=1$ would imply $2r+3s\ge 2$. Thus the offspring law is not deterministically one, and the extinction criterion applies. Extinction is certain.

For the specified probabilities, the fixed-point equation factors asIts roots in $[0,1]$ are $1/2$ and $1$, and the smaller root is the extinction probability of a branching process: