# Probabilty Mass Function. Function which depends on past outcomes $X$

I randomly draw numbers according to the probability mass function (PMF) $X$ in which all negative values have probability zero. Each value that is drawn from $X$ can be thought of the lifespan of one individual. I'd like to know what is the expected population size (number of existing individuals).

Note: We can assume that we are interested in the expected population size a long enough time after the start of the experiment so that the lifespan the individuals in the initial population can be safely ignored.

For example, I draw from $X$: [2,0,0,1,3,2,1,1]

Let's assume an initial population size of 0. If we call $Y$ the random variable (PMF) of the number of individuals in the population, then the observations according to the random variable $Y$ (population size) are [1,1,0,1,1,2,3,1]

Given the PMF of $X$: [0.25,0.25,0.15,0.1,0.05,0.1,0.1] for the values [0,1,2,3,4,5,6]. The cumulative PMF of $X$ is [0.25 0.50 0.65 0.75 0.80 0.90 1.00]

- What is expected value of $Y$ (the mean population size)?

- What is the PMF of $Y$ (the mean population size)?

Note: I know that having a population size of 0 at a given time step and a population size greater than zero at the following time step is not biologically realistic.

I didn't really know how to tackle this problem so I started with a very simple computer simulation hoping that I could find the solution from empirical data! I get lost because the function that maps $X$ to $Y$ does not seem to be a standard function but instead it depends on past outcomes of $X$

Here is the program (coded in R):

# Parameters
values = 0:6
proba = c(0.25,0.25,0.15,0.1,0.05,0.1,0.1)

CumulProba = cumsum(proba)
expected = 0
for (i in 1:length(values)){
expected = expected + values[i] * proba[i]
}

# Create observations from X
X = c()
for (i in 1:5000){
n = runif(1,0,1)
for (j in 1:length(CumulProba)){
if (n < CumulProba[j]){
X = append(X, values[j])
}
}
}

# Create the variable observation according to Y
Y = c()
for (i in 1:length(X)){
n = 0
for (j in 0:6){
if (i <= j){break}
if (X[i-j] > j){n = n + 1}
}
Y = append(Y, n)
}

# print some results
print(mean(Y))
print(mean(X))
print(expected)
print(mean(values))
print(var(X))
print(var(Y))

-
What was the output of the program? – jazzwhiz Feb 22 '14 at 2:18

We first note that $Y(t)$ only depends on $X(t-5)$ through $X(t)$. Then we have $$Y(t)=\sum_{i=0}^5 \begin{cases} 1\qquad&X(t-i)>i\\ 0&\text{else} \end{cases}$$ Then we have the following $P(X>5)=0.1$, $P(X>4)=0.2$, $P(X>3)=0.25$, $P(X>2)=0.35$, $P(X>1)=0.5$, and $P(X>0)=0.75$. Since the $X(t)$ are independent, $$E[Y(t>5)]=\sum_{i=0}^5P(X>i)=2.15$$ For early values just reduce the sum as appropriate. Hopefully this agrees with your computer program.