David Lewis

2021-12-30

I have a real valued number ${y}_{t}$. At each time step t, ${y}_{t}$ is multiplied by $\left(1+ϵ\right)$ with probability p and multiplied by $\left(1-ϵ\right)$ with probability $1-p$. What is the expected value of ${y}_{t+n}$? What is the variance?

encolatgehu

It is quite simple if you use independence in a more direct way and the method works for any distribution. Assuming that ${y}_{n}=\prod _{1\le k\le n}{X}_{k}$ where the ${X}_{k}$ are i.i.d. factors. By independence:
$\mathbb{E}{y}_{n}=\mathbb{E}\prod _{k}{X}_{k}=\prod _{k}\mathbb{E}{X}_{k}={\left(\mathbb{E}{X}_{k}\right)}^{n}$
valid for any distribution. In the specific (Bernoulli) example:
$\mathbb{E}{X}_{k}=\left(1+ϵ\right)p+\left(1-ϵ\right)\left(1-p\right)$. Similarly
$\mathbb{E}{y}_{n}^{2}=\mathbb{E}\prod _{k}{X}_{k}^{2}=\prod _{k}\mathbb{E}{X}_{k}^{2}={\left(\mathbb{E}{X}_{k}^{2}\right)}^{n}$
again valid for any distribution. In our case: $\mathbb{E}{X}_{k}^{2}={\left(1+ϵ\right)}^{2}p+{\left(1-ϵ\right)}^{2}\left(1-p\right)$. In particular, $var={\left(\mathbb{E}{X}_{k}^{2}\right)}^{n}-\left({\left(\mathbb{E}{X}_{k}\right)}^{2n}$ and you may carry on from there, in order to calculate limits etc... (e.g. gives a nice limit). The actual distribution of ${y}_{n}$ is in general quite complicated.

Lakisha Archer

Assuming independence and regarding the expectation:
For the sake of simplicity, let ${y}_{0}=1$. The value of our variable, at the ${n}^{th}$ moment, is
${\left(1+ϵ\right)}^{k}{\left(1-ϵ\right)}^{n-k}$
with probability
$\left(\begin{array}{c}n\\ k\end{array}\right){p}^{k}{\left(1-p\right)}^{n-k}$
(Where k denotes the number of multiplications by $1+ϵ$.)
The expected value is, then
$\sum _{k=0}^{n}{\left(1+ϵ\right)}^{k}{\left(1-ϵ\right)}^{n-k}\left(\begin{array}{c}n\\ k\end{array}\right){p}^{k}{\left(1-p\right)}^{n-k}=$
$=\sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right){\left[\left(1+ϵ\right)p\right]}^{k}{\left[\left(1-ϵ\right)\left(1-p\right)\right]}^{n-k}=$
$={\left[\left(1+ϵ\right)p+\left(1-ϵ\right)\left(1-p\right)\right]}^{n}={\left[1+2ϵp-ϵ\right]}^{n}$
because of the binomial theorem.
Unfortunately I could not find the trick for the calculation of the expectation of the square of the same random variable...

Vasquez

Following up zoli's answer, I think $\mathbb{E}\left({y}_{t}^{2}\right)$ can be found the same way, with the Binomial theorem yielding (assuming ${y}_{0}=1$)
$\mathbb{E}\left({y}_{t}^{2}\right)={y}_{0}^{2}\mathbb{E}\left(\left(1+ϵ{\right)}^{2}\left(1-ϵ{\right)}^{2}\right)$
$=\left[\left(1+ϵ{\right)}^{2}p+\left(1-ϵ{\right)}^{2}\left(1-p\right){\right]}^{n}$
$=\left[\left(1-ϵ{\right)}^{2}+4ϵp{\right]}^{n}$
So the variance is ${\sigma }_{n}^{2}=\left[\left(1-ϵ{\right)}^{2}+4ϵp{\right]}^{n}-\left[1+2ϵp-ϵ{\right]}^{2n}$. I don't know if it can be simplified any further.
Quick check: as

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