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There is a population that is equally likely to grow by $10\%$ or decrease by $10\%$ every day. What can we say about it in a year's time: will it be bigger, smaller, or the same as it was in the beginning? And in one year and one day?

Let the population size initially be equal to $x$. Then after first day we have in average:

$x\cdot \frac{1}{2} \cdot (1+0.1+1-0.1) = x$, which corresponds with initial value.

After second day average population size is equal:

$x\cdot \frac{1}{4} \cdot ((1+0.1)^2+2\cdot (1+0.1)\cdot (1-0.1) + (1-0.1)^2) = x$

Third step gives:

$\frac{1}{8} \cdot x \cdot((1+0.1)^3+3\cdot((1+0.1)^2) \cdot (1-0.1)+3 \cdot ((1-0.1)^2) \cdot (1+0.1)+(1-0.1)^3) = x$

However, when deriving the last formula, I separated the intermediate states up-down from down-up, etc, and final states such as up-up-down from up-down-up and down-up-up, etc. If this was not done, I received a different answer: the population was not stationary: $\frac{1}{6} \cdot x \cdot((1+0.1)^3+2\cdot((1+0.1)^2) \cdot (1-0.1)+2 \cdot ((1-0.1)^2) \cdot (1+0.1)+(1-0.1)^3) > x$

How is it correct to calculate the number of paths here: taking into account the order or without taking order into account? And is it true, if the order is taken into account, that I will always get the binomial formula and in the end the answer to the problem will be "The population does not change the number on average"?

Thank you

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    $\begingroup$ I edited your posting to correct (for example) $10%$, which should have been written $10\%$ so that the $~\%~$ symbol shows. $\endgroup$ Commented Apr 29 at 22:32
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    $\begingroup$ I think the 2's in your binomial expression should be 3's and the denominator should be 8, not 6. $\endgroup$ Commented Apr 29 at 22:59

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On an up day the population is multiplied by $1.1$. On a down day it is multiplied by $0.9$. As multiplication is commutative, it doesn't matter what order of days you have, just the number of ups and downs. After $m$ up days and $n$ down days the population is multiplied by $1.1^m\cdot 0.9^n$

You are correct that the expected value of the population does not change, but one up and one down results in multiplication by $0.99$. After a large number of days you expect to have about the same number of ups and downs, so the population is very likely to be smaller than when you started. The answer to this seeming conflict is that if you have rather more up days than down days the population is enormously larger than when you started. This is a low probability, true, but it pulls the average back up.

As an example, suppose we have a series of $100$ days. If you have $50$ up days and $50$ down days the population is down to about $0.605$ of what you started with. It takes $53$ up days and $47$ down days to be ahead and it is only $10\%$. But with $70$ ups and $30$ downs you are up a factor of $33.5$

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So this is a nice example of a situation when the expected value is not useful when determining the probability of a random variable being positive or negative.

In this case, we can compute that probability (almost) explicitly. As Ross Millikan mentioned, if $X_n$ is the population on day $n$ and $K_n$ is the number of days of population increase, then $$ X_n = (1.1)^{K_n} \cdot (0.9)^{n - K_n} = \biggl(\frac{11}{9}\biggr)^{K_n} \cdot \biggl(\frac{9}{10}\biggr)^{n} $$ so that \begin{align*} \Pr(X_k > 1) = \Pr\biggl(\biggl(\frac{11}{9}\biggr)^{K_n} \cdot \biggl(\frac{9}{10}\biggr)^{n} > 1\biggr) = \Pr\biggl(\frac{K_n}{n} > \frac{\log(10/9)}{\log(11/9)}\biggr). \end{align*}

But by the central limit theorem $\sqrt{n}(K_n/n - 1/2) \xrightarrow{\mathrm{d}} N(0, 1/4)$, so we can approximate \begin{align*} \Pr(X_k > 1) &= \Pr\biggl(\sqrt{n}\biggl(\frac{K_n}{n} - \frac{1}{2}\biggr) > \sqrt{n}\biggl(\frac{\log(10/9)}{\log(11/9)} - \frac{1}{2}\biggr)\biggr) \\ &\approx \Pr(Z/2 > 0.025\sqrt{n}), \end{align*} where $Z$ is a standard Gaussian.

But when $n = 365$, the above becomes $\Pr(Z > 0.05\sqrt{365}) \approx 0.17$. So despite the fact that $X_n$ has an expected value of $1$, it is greater then $1$ with a probability of only $17\%$. That is, the population will most likely decrease.

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    $\begingroup$ In fact it's even possible to construct an example where $E(X_n) > 1$ but $P(X_n > 1) \to 0$ as $n \to \infty$ - that is, the population most likely decreases over time but the expected value grows over time! For example, imagine a population that is equally likely to grow by 60% or shrink by 40% on any given day. $\endgroup$ Commented Apr 30 at 18:47

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