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Let's imagine, we start with one single bacterium. At each time step (generation), each bacterium has $x$ offspring and it dies (semelparous species). $x$ is a value drawn from a normal distribution with mean=$M$, standard deviation=$SD$.

Question 1:

What is the probability distribution of the number of bacteria (population size) after t generations ?

Question 2:

Same question but assuming that nobody ever dies ! So that after the very first reproductive event, there are $n$ bacteria, value which is drawn from a normal distribution mean = $M+1$, standard deviation = $SD$.

FastingGuy says the the probability distribution is normal with mean = $M*t$, standard deviation = $sqrt(t)*SD$.

Below is a very simple R-script that shows that the population size is almost always lower when $SD$ is higher and therefore the mean should depend of $SD$. What am I missunderstanding. In my code I'm using a uniform distribution to avoid a crash because all individuals reproduce equally at each generation. Is this the reason that higher $SD$ yields to lower population size ?

a=c()
for (i in 1:500){a[i]=runif(1,min=1.45,max=1.55)}

b=c()
for (i in 1:500){b[i]=runif(1,min=1,max=2)}

plot(cumprod(a),log='y',xlab='generation',ylab='number of inds')
points(cumprod(b),col='red')

a.fec=mean(a)
b.fec=mean(b)
a.fit=c()
b.fit=c()
for (i in 1:250){
    a.fit[i]=a[i*2]*a[(i*2)-1]
    b.fit[i]=b[i*2]*b[(i*2)-1]
}
a.fit=mean(a.fit)
b.fit=mean(b.fit)

enter image description here

Y is in logarithmic scale !

Thank you.

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  • $\begingroup$ Seems to be a Branching process. $\endgroup$ Commented Jul 26, 2013 at 13:00

2 Answers 2

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Let us ignore in this answer the odd feature that this model entails some negative populations of bacteria and let us start from the fact that the sum of $n$ i.i.d. normal random variables with mean $\mu$ and variance $\sigma^2$ has mean $n\mu$ and variance $n\sigma^2$. Hence, $E[X_{t+1}\mid X_t]=M\cdot X_t$ and, since $X_0=1$, $E[X_t]=M^t$ for every integer $t\geqslant0$.

Note however that $X_t$ is far from being normal when $t\geqslant2$. To wit, recall that any normal random variable $\xi$ with mean $\mu$ and variance $\sigma^2$ is such that $E[\mathrm e^{z\xi}]=\mathrm e^{z\mu+z^2\sigma^2/2}$ for every complex number $z$. Here, $E[\mathrm e^{zX_{t+1}}\mid X_t]=\mathrm e^{zMX_t+z^2\mathrm{SD}^2X_t/2}$, that is $E[\mathrm e^{zX_{t+1}}]=E[\mathrm e^{u(z)X_{t}}]$, where $u$ denotes the function defined by $u(z)=Mz+\frac12\mathrm{SD}^2z^2$. In particular, $X_1$ is normal since $u(z)$ is polynomial with degree $2$ but $X_t$ is not normal for $t\geqslant2$ since $u^{\circ t}(z)$ is polynomial but with degree $2^t\gt2$.

Likewise, $E[X_t]\ne M\cdot t$ in general and $\mathrm{var}(X_t)\ne \mathrm{SD}^2\cdot t$ in general.

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As mentioned, this seems to be a branching process. Let $Z_n$ be the the size of generation $n$ and $X_{n,i}$ the number of direct successors of member $i$ in generation $n$. Note that $X_{n,i}$ are iid normal random variables with mean $\mu$ and variance $\sigma^{2}$. Then $$Z_{n+1} = \sum_{i=1}^{Z_n} X_{n,i}$$

which is a sum of iid normal random variables.

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  • $\begingroup$ Does it mean that the mean of my new normal distribution $Zn$ is $M*t$ and the Standard deviation is $SD*t$. Is it correct ? $\endgroup$
    – Remi.b
    Commented Jul 26, 2013 at 13:14
  • $\begingroup$ The standard deviation would be $\sqrt{t} \cdot SD$. $\endgroup$ Commented Jul 26, 2013 at 13:18
  • $\begingroup$ And the mean is $M*t$ ? If so there's a result of a very simple simulation that I don't get. I'll add the R-script on the post $\endgroup$
    – Remi.b
    Commented Jul 26, 2013 at 13:20
  • $\begingroup$ yes is would be $\endgroup$ Commented Jul 26, 2013 at 13:22
  • $\begingroup$ PEV: Do you really think the distribution of the sum of a random number of i.i.d. normal random variables is normal? Your answer is designed to make us believe that, as well as your comments, but this is wrong. $\endgroup$
    – Did
    Commented Mar 2, 2015 at 6:23

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