HW 3

Exercise 12 An annual plant produces \(N\) seeds in a season which are assumed to have a Poisson distribution with parameter \(\lambda\). Each seed has a probability \(p\) of germinating to create a new plant which propagates in the following year. Let \(M\) be the random variable of the number of new plants.Then \[ P(M=m)=\frac{(p\lambda)^m}{m!}e^{-p\lambda} \qquad m=0,1,\ldots \] Find the mean number of plants in year \(k\). Show that the extinction is certain if \(p\lambda\le 1\).

Solution

The mean number of plants in year \(k\) is \[ E(X_k)=\mu^k =(p\lambda)^k. \] Since \(\mu=p\lambda\le 1\), it follows from the Extinction Probability Theorem that the extinction is certain, i.e., \(p_e=1\).

Exercise 13

Let \(X_1, X_2, \ldots\) be independent identically distributed random variables of which each can take the values -1 and +1 with probability 1/2. Show that the partial sums \[ Z_n=\sum_{j=1}^n \frac{1}{j} X_j \qquad n=1,2,\ldots \] form a martingale with respect to \(X_1, X_2, \ldots\).

Hint Show that \[ E\left( Z_n+\frac{1}{n+1}X_{n+1}\ |\ X_1, \ldots , X_n\right) =Z_n. \]

Solution

Observe that \[ E(X_i)=(-1)\frac{1}{2}+(1)\frac{1}{2}=0. \] Therefore, \[ E|Z_n|=E|\sum_{j=1}^n \frac{1}{j}X_j|=0<\infty. \] Furthermore, \[ E(Z_{n+1}\ |\ X_1, X_2, \ldots, X_n)=E(Z_{n}+\frac{1}{n+1}X_{n+1}\ |\ X_1, X_2, \ldots, X_n)= E(Z_n), \] where we used that \(Z_n\) and \(X_{n+1}\) are independent and \(E(X_{n+1})=0\). Therefore \(Z_n\) is a martingale because the two conditions in the definition hold.

Exercise 14

Let \(X_1, X_2, \ldots\) be independent random variables with means \(\mu_1, \mu_2, \ldots\), respectively. Let \[ Z_n= X_1+X_2+\ldots +X_n, \] and let \(Z_0=X_0=0\). Show that the random variable \[ Y_n=Z_n-\sum_{i=1}^n \mu_i \qquad n=1,2,\ldots \] is a martingale with respect to \(X_1, X_2, \ldots\).

Hint Note that \(E(Z_n\ |\ X_1, X_2, \ldots, X_n)=Z_n\).

Solution

Observe that \[ E|Y_n|=E|Z_n-\sum_{k=1}^n \mu_k|=E|\sum_{k=1}^n-\sum_{k=1}^n|=0<\infty. \] Additionally, \[ E(Y_{n+1}\ |\ X_1, X_2, \ldots, X_n)=E(Z_{n+1}-\sum_{k=1}^{n+1} \mu_k\ |\ X_1, X_2, \ldots, X_n) = E(Z_{n}+X_{n+1}-\sum_{k=1}^{n+1} \mu_k\ |\ X_1, X_2, \ldots, X_n)\\ =E(Z_{n}+\mu_{n+1}-\sum_{k=1}^n \mu_k-\mu_{n+1}\ |\ X_1, X_2, \ldots, X_n) = E(Z_{n}-\sum_{k=1}^n \mu_k\ |\ X_1, X_2, \ldots, X_n)=Y_n. \] Therefore, by definition, \(Y_n\) is a martingale.