We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$. Show that the sequence $(u_n)_{n\in\mathbb{N}}$ is increasing, then that it is convergent. We denote its limit by $l$.
We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$. Show that the equation $f(x)=x$ admits a smallest solution. In what follows, we denote it by $x_f$.
We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$. We denote by $x_f$ the smallest solution of $f(x)=x$ and by $l$ the limit of $(u_n)$. Show that $l=x_f$.
We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We set $m=f'(1)$. We denote by $x_f$ the smallest solution of $f(x)=x$. We assume $m>1$. Show that $x_f\in[0,1[$.
We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We set $m=f'(1)$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$. We denote by $x_f$ the smallest solution of $f(x)=x$. We now assume $m\leqslant 1$. Show that $x_f=1$ and that for all $n\in\mathbb{N}$, $u_n\neq 1$.
We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We set $m=f'(1)$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$. In this question, we assume $m=1$. We set, for $n\in\mathbb{N}$, $\varepsilon_n=1-u_n$. Show that $\lim_{n\rightarrow+\infty}\left(\frac{1}{\varepsilon_{n+1}}-\frac{1}{\varepsilon_n}\right)=\frac{f''(1)}{2}$.
We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We set $m=f'(1)$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$. In this question, we assume $m=1$. We set, for $n\in\mathbb{N}$, $\varepsilon_n=1-u_n$. Deduce that, as $n$ tends to infinity, $1-u_n\sim\frac{2}{f''(1)n}$. One may use Cesaro's lemma: if $(a_n)_{n\in\mathbb{N}}$ is a sequence of real numbers converging to $l$ and if we set, for $n\in\mathbb{N}^*$, $b_n=\frac{1}{n}(a_1+\cdots+a_n)$, then the sequence $(b_n)_{n\geqslant 1}$ converges to $l$.
We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We set $m=f'(1)$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$. We now assume $m<1$ and we set again, for $n\in\mathbb{N}$, $\varepsilon_n=1-u_n$. Show that the series with general term $\varepsilon_n$ is absolutely convergent and deduce the convergence of the series with general term $\ln\left(\frac{m^{-(n+1)}\varepsilon_{n+1}}{m^{-n}\varepsilon_n}\right)$.
We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We set $m=f'(1)$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$. We now assume $m<1$ and we set again, for $n\in\mathbb{N}$, $\varepsilon_n=1-u_n$. Deduce that there exists $c>0$ such that, as $n$ tends to infinity, $1-u_n\sim cm^n$.
Let $(X_n)_{n\in\mathbb{N}^*}$ be a sequence of random variables, mutually independent, with the same distribution taking values in $\mathbb{N}$, and let $T$ be a random variable taking values in $\mathbb{N}$ independent of the previous ones. We denote by $G_X$ the generating function common to all the $X_n$. For $n\in\mathbb{N}$ and $\omega\in\Omega$, we set $S_n(\omega)=\sum_{k=1}^n X_k(\omega)$ and $S_0(\omega)=0$, then $S(\omega)=S_{T(\omega)}(\omega)$. Show that, if $X$ and $Y$ are two independent random variables taking values in $\mathbb{N}$, then $G_{X+Y}=G_X G_Y$.
Let $(X_n)_{n\in\mathbb{N}^*}$ be a sequence of random variables, mutually independent, with the same distribution taking values in $\mathbb{N}$, and let $T$ be a random variable taking values in $\mathbb{N}$ independent of the previous ones. We denote by $G_X$ the generating function common to all the $X_n$. For $n\in\mathbb{N}$ and $\omega\in\Omega$, we set $S_n(\omega)=\sum_{k=1}^n X_k(\omega)$ and $S_0(\omega)=0$, then $S(\omega)=S_{T(\omega)}(\omega)$. By admitting that, for all $k\in\mathbb{N}$, $S_k$ is independent of $X_{k+1}$, prove that, for all $k\in\mathbb{N}$, $G_{S_k}=(G_X)^k$.
Let $(X_n)_{n\in\mathbb{N}^*}$ be a sequence of random variables, mutually independent, with the same distribution taking values in $\mathbb{N}$, and let $T$ be a random variable taking values in $\mathbb{N}$ independent of the previous ones. We denote by $G_X$ the generating function common to all the $X_n$. For $n\in\mathbb{N}$ and $\omega\in\Omega$, we set $S_n(\omega)=\sum_{k=1}^n X_k(\omega)$ and $S_0(\omega)=0$, then $S(\omega)=S_{T(\omega)}(\omega)$. By admitting that, for all $n\in\mathbb{N}$, $T$ and $S_n$ are independent, show that $$\forall t\in\left[0,1\left[,\forall K\in\mathbb{N}\quad G_S(t)=\sum_{k=0}^K P(T=k)\left(G_X(t)\right)^k+\sum_{n=0}^\infty\left(\sum_{k=K+1}^\infty P(T=k)P\left(S_k=n\right)t^n\right)$$
Let $(X_n)_{n\in\mathbb{N}^*}$ be a sequence of random variables, mutually independent, with the same distribution taking values in $\mathbb{N}$, and let $T$ be a random variable taking values in $\mathbb{N}$ independent of the previous ones. We denote by $G_X$ the generating function common to all the $X_n$. For $n\in\mathbb{N}$ and $\omega\in\Omega$, we set $S_n(\omega)=\sum_{k=1}^n X_k(\omega)$ and $S_0(\omega)=0$, then $S(\omega)=S_{T(\omega)}(\omega)$. For $K\in\mathbb{N}$ and $t\in\left[0,1\left[$, we set $R_K=\sum_{n=0}^\infty\left(\sum_{k=K+1}^\infty P(T=k)P\left(S_k=n\right)t^n\right)$. Show that $0\leqslant R_K\leqslant\frac{1}{1-t}\sum_{k=K+1}^\infty P(T=k)$.
Let $(X_n)_{n\in\mathbb{N}^*}$ be a sequence of random variables, mutually independent, with the same distribution taking values in $\mathbb{N}$, and let $T$ be a random variable taking values in $\mathbb{N}$ independent of the previous ones. We denote by $G_X$ the generating function common to all the $X_n$. For $n\in\mathbb{N}$ and $\omega\in\Omega$, we set $S_n(\omega)=\sum_{k=1}^n X_k(\omega)$ and $S_0(\omega)=0$, then $S(\omega)=S_{T(\omega)}(\omega)$. Conclude that $G_S=G_T\circ G_X$.
Let $(X_n)_{n\in\mathbb{N}^*}$ be a sequence of random variables, mutually independent, with the same distribution taking values in $\mathbb{N}$, and let $T$ be a random variable taking values in $\mathbb{N}$ independent of the previous ones. We denote by $G_X$ the generating function common to all the $X_n$. For $n\in\mathbb{N}$ and $\omega\in\Omega$, we set $S_n(\omega)=\sum_{k=1}^n X_k(\omega)$ and $S_0(\omega)=0$, then $S(\omega)=S_{T(\omega)}(\omega)$. We have shown $G_S=G_T\circ G_X$. Deduce that, if $T$ and the $X_n$ have finite expectation, then so does $S$ and $E(S)=E(T)E(X_1)$.
During a laying, an insect lays a random number of eggs following a Poisson distribution with parameter $\lambda>0$. Then, the probability that a given egg becomes a new insect is $\alpha\in]0,1[$. Recall the generating function of a random variable following a Poisson distribution with parameter $\lambda$.
During a laying, an insect lays a random number of eggs following a Poisson distribution with parameter $\lambda>0$. Then, the probability that a given egg becomes a new insect is $\alpha\in]0,1[$. Using the composition relation $G_S=G_T\circ G_X$, determine the distribution of the number of insects resulting from the laying.
Let $\mu$ be a probability distribution characterized by the sequence $(p_k)_{k\in\mathbb{N}}$ with $\sum_{k=0}^{+\infty}p_k=1$ and $p_0+p_1<1$. We define the Galton-Watson process with random variables $(X_{n,i})_{(n,i)\in\mathbb{N}\times\mathbb{N}^*}$ independent and all following distribution $\mu$, with $Y_0=1$ and $$\begin{cases} Y_{n+1}(\omega)=0 & \text{if }Y_n(\omega)=0\\ Y_{n+1}(\omega)=\sum_{i=1}^{Y_n(\omega)}X_{n,i}(\omega) & \text{if }Y_n(\omega)\neq 0\end{cases}$$ We denote by $f$ the generating function of $\mu$ and, for $n\in\mathbb{N}$, $\varphi_n$ the generating function of $Y_n$. We have $\varphi_0(t)=t$ for $t\in[0,1]$. Show that, for all $n\in\mathbb{N}$, $\varphi_{n+1}=\varphi_n\circ f$.
Let $\mu$ be a probability distribution characterized by the sequence $(p_k)_{k\in\mathbb{N}}$ with $\sum_{k=0}^{+\infty}p_k=1$ and $p_0+p_1<1$. We define the Galton-Watson process with $Y_0=1$ and the recurrence above. We assume that every random variable following distribution $\mu$ has expectation equal to $m$ and a variance. Express, for $n\in\mathbb{N}$, the expectation of $Y_n$ in terms of $m$ and $n$.
Let $\mu$ be a probability distribution characterized by the sequence $(p_k)_{k\in\mathbb{N}}$ with $\sum_{k=0}^{+\infty}p_k=1$ and $p_0+p_1<1$. We define the Galton-Watson process with $Y_0=1$ and the recurrence above. We denote by $f$ the generating function of $\mu$ and $\varphi_n$ the generating function of $Y_n$. a) Verify that the probability of extinction is equal to the limit of the sequence $(\varphi_n(0))_{n\geqslant 0}$. b) Verify that we can apply the results of Part I to the sequence $(\varphi_n(0))_{n\geqslant 0}$.
Let $\mu$ be a probability distribution characterized by the sequence $(p_k)_{k\in\mathbb{N}}$ with $\sum_{k=0}^{+\infty}p_k=1$ and $p_0+p_1<1$. We define the Galton-Watson process with $Y_0=1$ and the recurrence above. We denote by $f$ the generating function of $\mu$ and $m$ the expectation of $\mu$. If $m\leqslant 1$, show that the probability of extinction is equal to 1.
We consider the Galton-Watson process with extinction time $T$ defined by: $$\omega\in\Omega\quad\begin{cases}T(\omega)=\min\{n\in\mathbb{N}\mid Y_n(\omega)=0\} & \text{if there exists }n\in\mathbb{N}\text{ such that }Y_n(\omega)=0\\ T(\omega)=-1 & \text{otherwise}\end{cases}$$ We assume $m<1$. Verify that $T$ has a finite expectation.
We consider the Galton-Watson process with extinction time $T$. We assume $m<1$. a) Show that, for all integer $n$, $P(Y_n\geqslant 1)\leqslant m^n$. b) Show that $E(T)=\sum_{n=0}^{+\infty}P(T>n)$. c) Deduce an upper bound for $E(T)$.
We consider the Galton-Watson process. We assume $m\leqslant 1$. We denote, for $n\in\mathbb{N}^*$, $Z_n=1+\sum_{i=1}^n Y_i$ and $Z=1+\sum_{n=1}^{+\infty}Y_n$. We admit that $Z$ is a random variable defined on $\bigcup_{k\in\mathbb{N}}\{Y_k=0\}$. Show that $Z$ is defined on a set of probability 1.
We consider the Galton-Watson process. We assume $m\leqslant 1$. We denote, for $n\in\mathbb{N}^*$, $Z_n=1+\sum_{i=1}^n Y_i$ and $Z=1+\sum_{n=1}^{+\infty}Y_n$. a) Show that, for all $k\in\mathbb{N}$, $(P(Z_n\leqslant k))_{n\in\mathbb{N}^*}$ is a convergent sequence. Determine its limit. b) Deduce that, for all $k\in\mathbb{N}$, $(P(Z_n=k))_{n\in\mathbb{N}^*}$ converges to $P(Z=k)$. c) Show that, for all $s\in\left[0,1\left[$, all $n\in\mathbb{N}^*$ and $K\in\mathbb{N}$, $$\left|G_{Z_n}(s)-G_Z(s)\right|\leqslant\sum_{k=0}^K\left|P(Z_n=k)-P(Z=k)\right|+\frac{s^K}{1-s}$$ d) Deduce that the sequence of functions $(G_{Z_n})$ converges pointwise to $G_Z$ on $[0,1]$.
We consider the Galton-Watson process. We assume $m\leqslant 1$. We denote, for $n\in\mathbb{N}^*$, $Z_n=1+\sum_{i=1}^n Y_i$ and $Z=1+\sum_{n=1}^{+\infty}Y_n$. We denote by $f$ the generating function of $\mu$ and $m$ the expectation of $\mu$. a) Express $G_{Z_1}$ in terms of $f$. b) We admit that, for all natural integer $n$ greater than or equal to 2 and for all $s\in[0,1]$, $G_{Z_n}(s)=sf(G_{Z_{n-1}}(s))$. Deduce that, for all $s\in\left[0,1\left[$, $G_Z(s)=sf(G_Z(s))$. c) Show that $Z$ has finite expectation if and only if $m<1$. Calculate the expectation when this is the case.
We assume that, for all $k\in\mathbb{N}$, $p_k=\frac{1}{2^{k+1}}$. We have $\varphi_n(t)\neq 1$ for $t\in[0,1[$, so we can set $a_n(t)=\frac{1}{\varphi_n(t)-1}$. Show that, for $t\in\left[0,1\left[$, the sequence $(a_n(t))_{n\in\mathbb{N}}$ is arithmetic.
We assume that, for all $k\in\mathbb{N}$, $p_k=\frac{1}{2^{k+1}}$. Deduce that, for $t\in\left[0,1\left[$ and $n\in\mathbb{N}$, $\varphi_n(t)=\frac{n+(1-n)t}{1+n-nt}$.
We assume that, for all $k\in\mathbb{N}$, $p_k=\frac{1}{2^{k+1}}$. We have $\varphi_n(t)=\frac{n+(1-n)t}{1+n-nt}$. Express, for $(n,k)\in\mathbb{N}^2$, $P(Y_n=k)$ in terms of $n$ and $k$.
We assume that, for all $k\in\mathbb{N}$, $p_k=\frac{1}{2^{k+1}}$. We have $\varphi_n(t)=\frac{n+(1-n)t}{1+n-nt}$. The extinction time $T$ is defined as before. Express, in terms of $n\in\mathbb{N}^*$, the probability of the event $T>n$. Does the variable $T$ have a finite expectation?
We assume that, for all $k\in\mathbb{N}$, $p_k=\frac{1}{2^{k+1}}$. Express, for $s\in\left[0,1\left[$, $G_Z(s)$ in terms of $s$. Deduce the distribution of $Z$.
We assume $m>1$. We study a slightly different problem: $k$ being a fixed strictly positive integer, we assume that there are $k$ individuals in generation 0. We denote by $W_n$ the number of individuals in the $n$-th generation and define $u_n$ as the probability that the sequence $(W_n)_{n\in\mathbb{N}^*}$ takes the value $k$ for the first time at rank $n$: $$u_n=P\left((W_n=k)\cap\left(\bigcap_{i=1}^{n-1}(W_i\neq k)\right)\right)$$ For $n$ and $r$ non-zero natural integers, $u_n^{(r)}$ is the probability that the sequence $(W_n)_{n\in\mathbb{N}^*}$ takes the value $k$ for the $r$-th time at rank $n$. Verify that the series $\sum_{n\geqslant 1}u_n s^n$ and $\sum_{n\geqslant 1}u_n^{(r)}s^n$ converge when $s\in[-1,1]$.
We assume $m>1$. We study the Galton-Watson process starting with $k$ individuals in generation 0, with $W_n$ the number of individuals in generation $n$. Show that $P(W_1>k)>0$.
We assume $m>1$. We study the Galton-Watson process starting with $k$ individuals in generation 0, with $W_n$ the number of individuals in generation $n$. Show that the probability that the sequence $(W_n)_{n\in\mathbb{N}^*}$ does not take the value $k$ is non-zero; we denote this probability by $u$. One may study separately the cases $p_0=0$ and $p_0>0$.
We assume $m>1$. We study the Galton-Watson process starting with $k$ individuals in generation 0, with $W_n$ the number of individuals in generation $n$. We define $u_n$ and $u_n^{(r)}$ as above. Let $n\in\mathbb{N}^*$ and $r$ a natural integer greater than or equal to 2. Show the relation $$u_n^{(r)}=\sum_{i=1}^{n-1}u_i u_{n-i}^{(r-1)}$$
We assume $m>1$. We study the Galton-Watson process starting with $k$ individuals in generation 0. We define $u_n$, $u_n^{(r)}$, $U(s)=\sum_{n=1}^{+\infty}u_n s^n$ and $U_r(s)=\sum_{n=1}^{+\infty}u_n^{(r)}s^n$ for $s\in[-1,1]$. Deduce that, for every strictly positive integer $r$, $U_r=U^r$ ($U^r$ denotes $U\times U\times\cdots\times U$ $r$ times).
We assume $m>1$. We study the Galton-Watson process starting with $k$ individuals in generation 0, with $W_n$ the number of individuals in generation $n$. We define $u_n$, $u_n^{(r)}$, $U(s)$ and $U_r(s)$ as above, and $u$ is the probability that $(W_n)$ does not take the value $k$. Show that the probability that the sequence $(W_n)_{n\in\mathbb{N}^*}$ takes the value $k$ infinitely many times is zero.
We assume $m>1$. We study the Galton-Watson process with $Y_n$ the number of individuals in generation $n$ (starting from 1 individual). Show that the probability that the sequence $(Y_n)_{n\in\mathbb{N}^*}$ takes any fixed value $k$ infinitely many times is zero.
Let $(A_n)_{n\in\mathbb{N}}$ be a sequence of events all with probability 1. Show that $P\left(\bigcup_{n\in\mathbb{N}}\overline{A_n}\right)=0$. What can be deduced for $P\left(\bigcap_{n\in\mathbb{N}}A_n\right)$?
We assume $m>1$. Let $\alpha$ be the probability of extinction and $\beta$ be the probability that the sequence $(Y_n)$ diverges to infinity. Show that $\alpha+\beta=1$.