David Lewis

2021-12-30

I have a real valued number $y}_{t$ . At each time step t, $y}_{t$ is multiplied by $(1+\u03f5)$ with probability p and multiplied by $(1-\u03f5)$ with probability $1-p$ . What is the expected value of $y}_{t+n$ ? What is the variance?

encolatgehu

Beginner2021-12-31Added 27 answers

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:

$\mathbb{E}{X}_{k}=(1+\u03f5)p+(1-\u03f5)(1-p)$ . 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:$\mathbb{E}{X}_{k}^{2}={(1+\u03f5)}^{2}p+{(1-\u03f5)}^{2}(1-p)$ . In particular, $var={\left(\mathbb{E}{X}_{k}^{2}\right)}^{n}-({\left(\mathbb{E}{X}_{k}\right)}^{2n}$ and you may carry on from there, in order to calculate limits etc... (e.g. $n\to \mathrm{\infty},\text{}\u03f5n\to \lambda$ gives a nice limit). The actual distribution of $y}_{n$ is in general quite complicated.

valid for any distribution. In the specific (Bernoulli) example:

again valid for any distribution. In our case:

Lakisha Archer

Beginner2022-01-01Added 39 answers

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

$(1+\u03f5)}^{k}{(1-\u03f5)}^{n-k$

with probability

$\left(\begin{array}{c}n\\ k\end{array}\right){p}^{k}{(1-p)}^{n-k}$

(Where k denotes the number of multiplications by$1+\u03f5$ .)

The expected value is, then

$\sum _{k=0}^{n}{(1+\u03f5)}^{k}{(1-\u03f5)}^{n-k}\left(\begin{array}{c}n\\ k\end{array}\right){p}^{k}{(1-p)}^{n-k}=$

$=\sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right){\left[(1+\u03f5)p\right]}^{k}{\left[(1-\u03f5)(1-p)\right]}^{n-k}=$

$={[(1+\u03f5)p+(1-\u03f5)(1-p)]}^{n}={[1+2\u03f5p-\u03f5]}^{n}$

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...

For the sake of simplicity, let

with probability

(Where k denotes the number of multiplications by

The expected value is, then

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

Expert2022-01-09Added 669 answers

Following up zoli's answer, I think

So the variance is

Quick check: as

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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