Suppose that a species reproduces as follows: with probability p0, an organism has no offspring;

Suppose that a species reproduces as follows: with...

Key Concepts

Branching Process

A branching process is a mathematical model used to describe the growth of populations where individuals reproduce independently. In these processes, each individual in one generation produces a random number of offspring in the next generation according to a fixed probability distribution. This model is widely used to understand extinction and survival probabilities in species or particle systems.

Probability Generating Function

The probability generating function (PGF) is a tool that encodes the probabilities of different numbers of offspring into a power series. In the context of branching processes, the PGF is defined as f(x) = p0 + p1*x + p2*x^2 + ... . This function is crucial because it compactly represents the offspring distribution, and various properties (like the mean and variance) can be derived from it.

Fixed Point Equation

The extinction probability in a branching process is found as the smallest nonnegative solution to the equation f(x) = x. This fixed point equation arises because, in a branching process, the probability that the population eventually dies out is the same as the probability that all branches (produced by an initial individual) die out. Therefore, the extinction probability is self-consistent, satisfying the equation where the generating function equals its argument.

Mean Offspring

The mean number of offspring (also known as the reproductive rate) is a key parameter in branching processes, calculated as the expected value using the offspring probabilities. It determines the process’s behavior: if the mean is greater than 1, the process is supercritical and there is a chance of survival (extinction probability less than 1); if it is equal to or less than 1, the process is critical or subcritical and extinction is almost sure (extinction probability equals 1).

Extinction Probability

The extinction probability is the chance that the population eventually dies out. In a supercritical branching process (mean > 1), there exists a fixed point solution of the form 0 < x < 1, meaning there is a positive probability for the species to survive. In contrast, in a critical or subcritical process (mean <= 1), the only nonnegative solution to the fixed point equation is x = 1, indicating that the species will almost surely go extinct.

Transcript

00:01 In problem 59 we want to show that the first equation here gives a smaller solution or a smaller probability of extension of extinction then the second equation here the solution of this equation or the smallest non -negative solution of these equations is the probability of extinction then we have first to solve the first equation or to get the root of the first equation and the root of the second equation to get the probability of extinction using the first and the second equation let's start by the first equation we can get the root using newton's method newitance method dependence on iterations based on an initial guess the second iteration x n plus 1 equals the previous iteration x n minus f of x n divided by f dash of x n we can evaluate or get f of x as the left -hand side of the equation only if the right -hand side equals 0 this means we have f of x here equals the left -hand side after we move x to the left -hand side then f of x equals 0 .1 plus 0 .2x plus 0 .3 x squared plus 0 .4 x cubed minus x then we can evaluate f dash of x equals 0 .2 plus 0 .6 x plus 1 .2 x squared minus 1.

02:07 Now we can use newton's method based on an initial guess we see that it's something of a probability then the root is about from 0 to 1 and we want to get the smallest most non -positive solution the smallest non -negative solution then we can start by x node equal 0 this mean we can use newton's method to get x1 equals x node minus f of x node divided by by f dash of x note to make our calculations clear we can make a table for all our iterations number of iterations initial value x equals 0 f of x we can get if of x from here and f dash of x we can get f dash of x from here f of x equals 0 .1 equals 0 .1 when you substitute x equals 0 f dash of 0 equals negative 0 .8 and applying newton's method equation we can get x1 equals 0 minus 0 .1 divided by negative 0 .8 then the x1 equals 0 .125 we can stop after we see that x2 or x n plus 1 equals x n we can see that x1 is not equal to x0 or x node then we continue iterations we get f of x we get f of x 1 when we substitute by x equals 1 quarter we get f of x1 equals 0 .0 .055 we get f dash equals minus 0 .7 or 6 applying newton's method equation we can get x2 equals 0 .1327 we can see that x2 is not equal to x1 for some sort of accuracy one decimal plus or two decimal place for example then we continue iterations we get f of x2 f of x2 equals 2 .717 multiplied by 10 to the power of negative 5 and if the of x equals minus 0 .6992 we use newton's method to get x3 equals 0 .1328 almost the same as x2 we can stop here and say that we have the root for the first equation x equals 0 .1328 and this is the probability of extinction to compare with the second equation, we repeat the previous steps.

05:36 We define f of x equals the left hand side of the second equation.

05:43 In case the right hand side equal 0, this means we move x here.

05:49 We have 0 .4 plus 0 .3x plus 0 .2 x squared plus 0 .1 x cubed minus x squared, plus 0 .2 x squared plus 0 .1 x cubed, minus x.

06:02 Then we can evaluate f dash of x equals 0 .3 plus 0 .4 x plus 0 .3 x squared minus 1 now we can use newton's method using the same initial guess because we talk about probabilities and probability starts from 0 to 1 and we want the smallest non -negative root then we start by x node equal zero and we make the same table for our calculations number of iteration number 0 x we have x equal 0 f of x and f dash of x equals 0 f of x equals 0 .4 and f of x node or f of 0 f of 0 f dash of x node or f dash of 0 equals negative 0 .7 using the nuitance method we can get x1 using the same general equation here.

07:20 We use this equation x n plus 1 equal x n minus f of xn divided by f dash of xn then x1 equals 0 0 .4 divided by negative 0 .7 which means we have x1 equals 0 .577114 we can see that x1 is not equal to x node we continue iterations we get f of x1 f of x1 equals 0 .083965 f dash of x1 equals minus 0 .3373 4 7 using the netherance method equation we get x2 equals 0 .796253 we continue iterations as long as x2 is not equals to x1 f of x2 equals to x2 equals x2 equals x2 equals f of x2 equals 0 .01911 if dash of x2 equals minus 0 .11913 can evaluate x3 using the method equation x3 equals 0 .9034 we continue iterations we get f of x3 0 .0087 if dash of x3 equals minus 0 .909 6 .6 6 8 using nettance method equation we get x4 equals 0 .9507 we continue iterations f of x4 equals 0 .0012 f dash of x2 equals minus 0 .0486 we use newton's method equation to get x5 x5 equals 0 .9755 we continue iterations f of x5 equals 0 .003 f dash of x5 equals minus 0 .244 using the nuisance method we get x6 x6 equals 0 .9776 we continue iterations we get f of x6 x6 equals 7 .4 7 x multiplied by 10 to the power of negative 5 we get f dash of x6 equals minus 0 .0 .0211...

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