I have a real valued number y_t. At each time

David Lewis

David Lewis

Answered question


I have a real valued number yt. At each time step t, yt is multiplied by (1+ϵ) with probability p and multiplied by (1ϵ) with probability 1p. What is the expected value of yt+n? What is the variance?

Answer & Explanation



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 yn=1knXk where the Xk are i.i.d. factors. By independence:
valid for any distribution. In the specific (Bernoulli) example:
EXk=(1+ϵ)p+(1ϵ)(1p). Similarly
again valid for any distribution. In our case: EXk2=(1+ϵ)2p+(1ϵ)2(1p). In particular, var=(EXk2)n((EXk)2n and you may carry on from there, in order to calculate limits etc... (e.g. n, ϵnλ gives a nice limit). The actual distribution of yn is in general quite complicated.
Lakisha Archer

Lakisha Archer

Beginner2022-01-01Added 39 answers

Assuming independence and regarding the expectation:
For the sake of simplicity, let y0=1. The value of our variable, at the nth moment, is
with probability
(Where k denotes the number of multiplications by 1+ϵ.)
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...


Expert2022-01-09Added 669 answers

Following up zoli's answer, I think E(yt2) can be found the same way, with the Binomial theorem yielding (assuming y0=1)
So the variance is σn2=[(1ϵ)2+4ϵp]n[1+2ϵpϵ]2n. I don't know if it can be simplified any further.
Quick check: as ϵ0, σn0

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