[Math] Rademacher random variables in terms of Bernoulli

probabilityprobability theoryrandomrandom variablesstatistics

I've found out that Rademacher random variables and Bernoulli random variables plays an important role in Probability theory. I am wondering how they are connected. For example,

Let $r_i, i=1, \ldots, n$ be Rademacher random variables, such that $P(r_i=1)=a$ and $P(r_i=-1)=1-a$.

Let $b_i, i=1, \ldots, n$ be Bernoulli random variables, such that $P(b_i=1)=b$ and $P(b_i=0)=1-b$.

Let $x_i, i=1, \ldots, n$ be real numbers.

Consider $A=\sum_{i=1}^nx_ir_i$.

How to represent $A$ in terms of $B=\sum_{i=1}^nx_ib_i$? What is the relation between $a$ and $b$?

Thank you.

Best Answer

Hints:

If $r_i$ is a Rademacher random variable, then consider $\dfrac{r_i+1}{2}$.

Now consider $\displaystyle\sum_i x_i \dfrac{r_i+1}{2}$ or equivalently $\dfrac{\displaystyle\sum_i x_i r_i}{2} + \dfrac{\displaystyle\sum_i x_i }{2}$.

Related Solutions

[Math] Average absolute value of sum with Rademacher random variables

We don't need to estimate it ... we can find the exact solution.

If $Z$~Bernoulli($\frac12$), then $X=2Z-1$ has a Rademacher distribution where $X=-1$ or $1$ with equal probability. Let $X_1, ..., X_n$ denote indepedent Rademacher random variables. Then, the pmf of:

$$ \sum_{i=1}^n X_i = (2 \sum_{i=1}^n Z_i) - n = 2 Y -n$$

where $Y$~ $Binomial(n,\frac12)$ (since the sum of $n$ Bernoulli's is $Binomial(n,p)$).

Let random variable $Y$ have pmf $f(y)$:

Then, the expectation we seek, namely, $E[ \left| \sum_{i=1}^n X_i \right|]$ can be found as $E[ \left| 2 Y - n \right|]$, derived here using mathStatica (of which I am one of the authors):

All done.

Here is a plot of the solution as $n$ increases from 1 to 100:

Monte Carlo check

When using computer algebra systems (or indeed working manually), it is always good practise to check one's work using alternative methods. For example, when $n = 130$, the exact solution derived above yields:

sol /. n -> 130.

9.07981

Here is a quick Monte Carlo 'check' in Mathematica ... when $n = 130$, we generate 100000 samples, each containing the sum of 130 Rademacher random variables, sum them up, take the absolute value, and compute the sample mean:

Mean[Table[ 
              Abs[Total[Table[RandomChoice[{-1, 1}], {130}]]], 
           {100000}]] // N

9.07358

Sum of Normalized Rademacher Random Variables

The law of the iterated logarithm states that \begin{align} \limsup_n\frac{S_n}{\sqrt{n}\log\log n}\to1 \text{ a.s.},\\ \liminf_n\frac{S_n}{\sqrt{n}\log\log n}\to-1 \text{ a.s.}, \end{align} where $S_n=\sum_1^nX_k$. In words, for each $\omega$ in a set of probability 1, the sample path $S_n(\omega)/(\sqrt{n}\log\log n)$ is arbitrary close to $1$ infinitely often and to $-1$ i.o. as $n\to\infty$. Therefore, $n^\epsilon\log\log n\to\infty$ for $\epsilon\ge 0$ implies that \begin{align} \limsup_n (n^\epsilon\log\log n) S_n(\omega)/(\sqrt{n}\log\log n)=\limsup_nS_n(\omega)/n^{1/2-\epsilon}=\infty\\ \liminf_n (n^\epsilon\log\log n) S_n(\omega)/(\sqrt{n}\log\log n)=\liminf_n S_n(\omega)/n^{1/2-\epsilon}=-\infty. \end{align} So for $p\le 1/2$ the process fails to converge almost surely.

Best Answer

Related Solutions

[Math] Average absolute value of sum with Rademacher random variables

Sum of Normalized Rademacher Random Variables

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