Yes, $T(a,b)$ has continuum cardinality when $a+b>1$... with positive probability. Of course there is also positive probability that $T(a,b)$ has no infinite paths, if $a<1$ and $b<1$.

$T(1,0)$ has only one infinite path, ao ``path'' means finite-or-infinite path.

#Background#
The relevant result, the Extinction Criterion below, was first stated by Bienaymé in 1845; see Heyde and Seneta (I. J. Bienaymé. Statistical theory anticipated) pp. 116--120 and Lyons and Peres (Probability on Trees and Networks, Cambridge University Press, in preparation. Current version available at [http://mypage.iu.edu/~rdlyons/][1]), Proposition 5.4. The first published proof of Bienaymé's theorem appears in Cournot (1847, De l'Origine et des Limites de la Correspondance entre l'Algèbre et la Géométrie, Hachette, Paris) pp. 83--86.
#Extinction Criterion#
Given numbers $p_{k}\in [0,1]$ with $p_{1}\ne 1$ and $\sum_{k\ge 0}p_{k}=1$, let $Z_{0}=1$, let $L$ be a random variable with $\mathbb P(L=k)=p_{k}$, let

$ \{ L^{(n)}_i \},\quad n,i\ge 1 $

be independent copies of $L$, and let $$ Z_{n+1}=\sum_{i=1}^{Z_{n}}L_{i}^{(n+1)}. $$ Let $q=\mathbb P((\exists n)\, Z_{n}=0)$. Then $q=1$ iff $\mathbb E(L)=\sum_{k\ge 0}kp_{k} \le 1$. Moreover, $q$ is the smallest fixed point of $f(s)=\sum_{k\ge 0}p_{k}s^{k}$.

Solution sketch

As stated the Extinction Criterion only shows that with positive probability there is at least one infinite path. Now consider, along this path, the possibility of branching off to create another infinite path. You get infinitely many independent events, so the probability that they all fail to happen is 0. Continuing this way, we get a positive probability of continuum many infinite paths.

Yes, $T(a,b)$ has continuum cardinality when $a+b>1$... with positive probability. Of course there is also positive probability that $T(a,b)$ has no infinite paths, if $a<1$ and $b<1$.

$T(1,0)$ has only one infinite path, so ``path'' means finite-or-infinite path.

#Background#
The relevant result, the Extinction Criterion below, was first stated by Bienaymé in 1845; see Heyde and Seneta (I. J. Bienaymé. Statistical theory anticipated) pp. 116--120 and Lyons and Peres (Probability on Trees and Networks, Cambridge University Press, in preparation. Current version available at [http://mypage.iu.edu/~rdlyons/][1]), Proposition 5.4. The first published proof of Bienaymé's theorem appears in Cournot (1847, De l'Origine et des Limites de la Correspondance entre l'Algèbre et la Géométrie, Hachette, Paris) pp. 83--86.
#Extinction Criterion#
Given numbers $p_{k}\in [0,1]$ with $p_{1}\ne 1$ and $\sum_{k\ge 0}p_{k}=1$, let $Z_{0}=1$, let $L$ be a random variable with $\mathbb P(L=k)=p_{k}$, let

$ \{ L^{(n)}_i \},\quad n,i\ge 1 $

be independent copies of $L$, and let $$ Z_{n+1}=\sum_{i=1}^{Z_{n}}L_{i}^{(n+1)}. $$ Let $q=\mathbb P((\exists n)\, Z_{n}=0)$. Then $q=1$ iff $\mathbb E(L)=\sum_{k\ge 0}kp_{k} \le 1$. Moreover, $q$ is the smallest fixed point of $f(s)=\sum_{k\ge 0}p_{k}s^{k}$.

Solution sketch

As stated the Extinction Criterion only shows that with positive probability there is at least one infinite path. Now consider, along this path, the possibility of branching off to create another infinite path. You get infinitely many independent events, so the probability that they all fail to happen is 0. Continuing this way, we get a positive probability of continuum many infinite paths.

Yes, $T(a,b)$ has continuum cardinality when $a+b>1$... with positive probability. Of course there is also positive probability that $T(a,b)$ has no infinite paths, if $a<1$ and $b<1$.

(By the way $T(1,0)$ has only one infinite path, by ``path'' do you mean finite-or-infinite path?)

The relevant result, the Extinction Criterion below, was first stated by Bienaymé in 1845; see Heyde and Seneta (I. J. Bienaymé. Statistical theory anticipated) pp. 116--120 and Lyons and Peres (Probability on Trees and Networks, Cambridge University Press, in preparation. Current version available at http://mypage.iu.edu/~rdlyons/), Proposition 5.4. The first published proof of Bienaymé's theorem appears in Cournot (1847, De l'Origine et des Limites de la Correspondance entre l'Algèbre et la Géométrie, Hachette, Paris) pp. 83--86.

Extinction Criterion: Given numbers $p_{k}\in [0,1]$ with $p_{1}\ne 1$ and $\sum_{k\ge 0}p_{k}=1$, let $Z_{0}=1$, let $L$ be a random variable with $\mathbb P(L=k)=p_{k}$, let

$ \{ L^{(n)}_i \},\quad n,i\ge 1 $

be independent copies of $L$, and let $$ Z_{n+1}=\sum_{i=1}^{Z_{n}}L_{i}^{(n+1)}. $$ Let $q=\mathbb P((\exists n)\, Z_{n}=0)$. Then $q=1$ iff $\mathbb E(L)=\sum_{k\ge 0}kp_{k} \le 1$. Moreover, $q$ is the smallest fixed point of $f(s)=\sum_{k\ge 0}p_{k}s^{k}$.

Well, as stated that only shows that with positive probability there is at least one infinite path. Now consider, along this path, the possibility of branching off to create another infinite path. You get infinitely many independent events, so the probability that they all fail to happen is 0. Continuing this way, we get a positive probability of continuum many infinite paths.

Yes, $T(a,b)$ has continuum cardinality when $a+b>1$... with positive probability. Of course there is also positive probability that $T(a,b)$ has no infinite paths, if $a<1$ and $b<1$.

$T(1,0)$ has only one infinite path, ao ``path'' means finite-or-infinite path.

#Background#
The relevant result, the Extinction Criterion below, was first stated by Bienaymé in 1845; see Heyde and Seneta (I. J. Bienaymé. Statistical theory anticipated) pp. 116--120 and Lyons and Peres (Probability on Trees and Networks, Cambridge University Press, in preparation. Current version available at [http://mypage.iu.edu/~rdlyons/][1]), Proposition 5.4. The first published proof of Bienaymé's theorem appears in Cournot (1847, De l'Origine et des Limites de la Correspondance entre l'Algèbre et la Géométrie, Hachette, Paris) pp. 83--86.
#Extinction Criterion#
Given numbers $p_{k}\in [0,1]$ with $p_{1}\ne 1$ and $\sum_{k\ge 0}p_{k}=1$, let $Z_{0}=1$, let $L$ be a random variable with $\mathbb P(L=k)=p_{k}$, let

$ \{ L^{(n)}_i \},\quad n,i\ge 1 $

be independent copies of $L$, and let $$ Z_{n+1}=\sum_{i=1}^{Z_{n}}L_{i}^{(n+1)}. $$ Let $q=\mathbb P((\exists n)\, Z_{n}=0)$. Then $q=1$ iff $\mathbb E(L)=\sum_{k\ge 0}kp_{k} \le 1$. Moreover, $q$ is the smallest fixed point of $f(s)=\sum_{k\ge 0}p_{k}s^{k}$.

Solution sketch

As stated the Extinction Criterion only shows that with positive probability there is at least one infinite path. Now consider, along this path, the possibility of branching off to create another infinite path. You get infinitely many independent events, so the probability that they all fail to happen is 0. Continuing this way, we get a positive probability of continuum many infinite paths.

Yes, $T(a,b)$ has continuum cardinality when $a+b>1$... with positive probability. Of course there is also positive probability that $T(a,b)$ has no infinite paths, if $a<1$ and $b<1$.

(By the way $T(1,0)$ has only one infinite path, by ``path'' do you mean finite-or-infinite path?)

The relevant result, the Extinction Criterion below, was first stated by Bienaymé in 1845; see Heyde and Seneta (I. J. Bienaymé. Statistical theory anticipated) pp. 116--120 and Lyons and Peres (Probability on Trees and Networks, Cambridge University Press, in preparation. Current version available at http://mypage.iu.edu/~rdlyons/), Proposition 5.4. The first published proof of Bienaymé's theorem appears in Cournot (1847, De l'Origine et des Limites de la Correspondance entre l'Alg`ebre et la G'eom'etrie, Hachette, Paris) pp. 83--86.

Extinction Criterion: Given numbers $p_{k}\in [0,1]$ with $p_{1}\ne 1$ and $\sum_{k\ge 0}p_{k}=1$, let $Z_{0}=1$, let $L$ be a random variable with $\mathbb P(L=k)=p_{k}$, let

$ \{ L^{(n)}_i \},\quad n,i\ge 1 $

be independent copies of $L$, and let $$ Z_{n+1}=\sum_{i=1}^{Z_{n}}L_{i}^{(n+1)}. $$ Let $q=\mathbb P((\exists n)\, Z_{n}=0)$. Then $q=1$ iff $\mathbb E(L)=\sum_{k\ge 0}kp_{k} \le 1$. Moreover, $q$ is the smallest fixed point of $f(s)=\sum_{k\ge 0}p_{k}s^{k}$.

Well, as stated that only shows that with positive probability there is at least one infinite path. Now consider, along this path, the possibility of branching off to create another infinite path. You get infinitely many independent events, so the probability that they all fail to happen is 0. Continuing this way, we get a positive probability of continuum many infinite paths.

Yes, $T(a,b)$ has continuum cardinality when $a+b>1$... with positive probability. Of course there is also positive probability that $T(a,b)$ has no infinite paths, if $a<1$ and $b<1$.

(By the way $T(1,0)$ has only one infinite path, by ``path'' do you mean finite-or-infinite path?)

The relevant result, the Extinction Criterion below, was first stated by Bienaymé in 1845; see Heyde and Seneta (I. J. Bienaymé. Statistical theory anticipated) pp. 116--120 and Lyons and Peres (Probability on Trees and Networks, Cambridge University Press, in preparation. Current version available at http://mypage.iu.edu/~rdlyons/), Proposition 5.4. The first published proof of Bienaymé's theorem appears in Cournot (1847, De l'Origine et des Limites de la Correspondance entre l'Algèbre et la Géométrie, Hachette, Paris) pp. 83--86.

Extinction Criterion: Given numbers $p_{k}\in [0,1]$ with $p_{1}\ne 1$ and $\sum_{k\ge 0}p_{k}=1$, let $Z_{0}=1$, let $L$ be a random variable with $\mathbb P(L=k)=p_{k}$, let

$ \{ L^{(n)}_i \},\quad n,i\ge 1 $

be independent copies of $L$, and let $$ Z_{n+1}=\sum_{i=1}^{Z_{n}}L_{i}^{(n+1)}. $$ Let $q=\mathbb P((\exists n)\, Z_{n}=0)$. Then $q=1$ iff $\mathbb E(L)=\sum_{k\ge 0}kp_{k} \le 1$. Moreover, $q$ is the smallest fixed point of $f(s)=\sum_{k\ge 0}p_{k}s^{k}$.

Well, as stated that only shows that with positive probability there is at least one infinite path. Now consider, along this path, the possibility of branching off to create another infinite path. You get infinitely many independent events, so the probability that they all fail to happen is 0. Continuing this way, we get a positive probability of continuum many infinite paths.