-1
$\begingroup$

Consider this distirbution,

$S=a_0\cdot b_0 + a_1\cdot b_1 + a_2\cdot b_2 + \cdots +a_{n-1}\cdot b_{n-1}$

Now, I want find the expected value of $S$ when, $a_i$'s are selected from $\{0,1\}$ with equal probability. And $b_i$'s are uniformly selected from $[-1/2,1/2]$. All values are i.i.d.

According to my calculation, $\mu_a=0\ \&\ \mu_b=0$, $E[a^2]=1/2\ \&\ E[b^2]=(b-a)^2/12=1/12$

As mean is zero in the combined distribution, the variance of the joint distribution is $E[a^2\cdot b^2]=E[a^2]E[b^2]=1/24$. Hence, the variance of $S=n/24$ and $\sigma=\sqrt{n/24}$.

Now, in this paper (page 9), the authors say that

Each entry of the vector $Ws$ is a sum of $n$ (or around $n/2$ in the case where $s \in \{0, 1\}^n$) rational numbers in the interval $[−1/2, 1/2]$. Assuming the entries of $W$ are uniformly distributed then the central limit theorem suggests that each entry of $Ws$ has an absolute value roughly $1/4\sqrt{n/2}$.

I did not understand how the authors arrived at this value.

It is to be noted that the values of $W$ are sampled uniformly from $[-1/2,1/2]$ similar to the $b_i$'s and the values of $s$ is sampled from $\{0,1\}$ with equal probability similar to the $a_i$'s.

Also, In section 2.1 the authors say that the expected value (Euclidean norm) of a $n$ dimensional vector sampled from a discrete Gaussian distribution with s.d $\sigma$ is $\sqrt{n}\sigma$. So, according to the authors, $S$ is a summation $n/2$ values ($b_i$) sampled uniformly from $[−1/2, 1/2]$, so can we say that central limit theorem guarantees that the distribution of $b_i$ follow a Gaussian distribution with $\sigma=1/4$? I appreciate your help.

EDIT : $\mu_a$ will be $1/2$ and not $0$. But it still does not solve the issue. as $\mu_a \cdot \mu_b=0$

$\endgroup$
  • $\begingroup$ The general idea of using the CLT seems OK, but some of your statements puzzle me: (a) Isn't $E(a_i) = \mu_a = E(a_i^2) = 1/2(0) + 1/2(1) = 1/2?$ (b) What are the entries of $W.$ (It's unfair to ask me to read the whole paper, which presumably you have done.) (c) Am I correct that the $a_i$'s are symmetrical Bernoulli's (discrete) and that $b_i \stackrel{iid}{\sim} \mathsf{Unif}(-.5,.5)$ (continuous)? (d) What do you mean by the 'value' of a vector? $\endgroup$ – BruceET Jul 13 '17 at 23:48
  • $\begingroup$ @BruceET I thank you for your time and I apologize for the confusions. I will clarify them here, a) Yes you are right, I miscalculated the mean, b) the entries of $W$ is uniformly sampled from $[-/5,.5]$ c)You are right about $a_i$, each entry of $a_i$ can assume $0$ or $1$ with probability $1/2$. $b_i$ is also correct. d) by the value of a vector I mean the Euclidean norm. $|| \cdot ||_2$. I have made corrections in question. $\endgroup$ – Rick Jul 14 '17 at 8:16
  • $\begingroup$ So, it seems clear that $SD(S) = \sigma = \sqrt{n/24}.$ I suppose your edit is an improvement, but exactly what remains unresolved? $W$ is not $S.$ $\endgroup$ – BruceET Jul 14 '17 at 17:55
  • $\begingroup$ @BruceET I don't understand authors arrived at the different value from mine (the paper was accepted in a reputed peer reviewed conference). I don't understand who is right? $\endgroup$ – Rick Jul 15 '17 at 8:59
  • $\begingroup$ You say "...according to the authors, $S$ is a summation $n/2$ values ($b_i$) sampled uniformly from $[−1/2, 1/2]$," In fact, on average $S$ is a summation of $n/2$ of such uniforms. I don't know which is wrong, their statement or your interpretation of it. You know the right answer to the the Question you pose (variance of $S$). Suggest you read the paper more carefully to make sure of your interpretation of $W.$ And if the authors or wrong, it would hardly be the first such instance in a 'reputable refereed journal'.. $\endgroup$ – BruceET Jul 15 '17 at 20:27
1
$\begingroup$

Comment continued: Assuming my guesses in (c) of my Comment are correct, here is a simulation of a million sums $S$ for $n = 20.$ By part (a) of my Comment, you should have $E(S) = 0.$ Also, you say you should have $\sigma = \sqrt{20/24} = 0.9128709,$ which is approximately confirmed by the simulation. For large $n,$ you will have $S$ nearly (but not exactly) normal. It seems that $n = 20$ is large enough to get a good approximation to normality.

m = 10^6;  s = numeric(m);  n = 20
for(i in 1:m) {
  a = rbinom(n, 1, .5);  b = runif(n, -.5,.5)
  s[i] = sum(a*b) }
mean(s);  sd(s)
## -0.0004134797    # aprx E(S) = 0 
## 0.9128541        # aprx SD(S) = 0.9128709

The histogram below shows the approximate distribution of $S$ from simulation together with the well-fitting density function of $\mathsf{Norm}(0, .9129).$

enter image description here

Of course, the simulation doesn't 'prove' anything, but its results may encourage you that you are mainly on the right track. You should be able to clarify your statement of the problem and write a convincing solution for general $n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.