Proving the Uniform Distribution of Y from Independent Random Variables X

[bennyzadir]

Let be $X_1, X_2, ..., X_n, ... $ independent identically distributed random variables with mutual distribution $ \mathbb{P}\{X_i=0\}=1-\mathbb{P}\{X_i=1\}=p $. Let be $ Y:= \sum_{n=1}^{\infty}2^{-n}X_n$.
a) Prove that if $p=\frac{1}{2}$ then Y is uniformly distributed on interval [0,1].
b) Show that if $p \neq \frac{1}{2}$ then the distribution function of random variable Y is continuous but not absolutely continuous and it is singular (i.e. singular with respect to the Lebesque measure, i.e with respect to the uniform distribution).

I would really appreciate if you could help me!
Thank you in advance!

 

[chisigma]

Let be $X_1, X_2, ..., X_n, ... $ independent identically distributed random variables with mutual distribution $ \mathbb{P}\{X_i=0\}=1-\mathbb{P}\{X_i=1\}=p $. Let be $ Y:= \sum_{n=1}^{\infty}2^{-n}X_n$.
a) Prove that if $p=\frac{1}{2}$ then Y is uniformly distributed on interval [0,1].
b) Show that if $p \neq \frac{1}{2}$ then the distribution function of random variable Y is continuous but not absolutely continuous and it is singular (i.e. singular with respect to the Lebesque measure, i.e with respect to the uniform distribution).
I would really appreciate if you could help me!
Thank you in advance!

If You set $\varphi_{n}(x)$ the p.d.f. of each $X_{n}$ and set $\Phi_{n}(\omega)=\mathcal {F} \{\varphi_{n}(x)\}$ You have that the p.d.f. of $\displaystyle Y=\sum_{n=1}^{\infty} 2^{-n}\ X_{n}$ is...

$\displaystyle \Phi(\omega)= \prod_{n=1}^{\infty} \Phi_{n}(\omega)$ (1)

If $p=\frac{1}{2}$ is...

$\displaystyle \varphi_{n} (x)= \frac{1}{2}\ \delta(x) + \frac{1}{2}\ \delta(x-\frac{1}{2^{n}}) \implies \Phi_{n}(\omega)= e^{- i \frac{\omega}{2^{n+1}}}\ \cos \frac {\omega}{2^{n}}$ (2)

Now You have to remember that is...

$\displaystyle \frac{\sin \omega}{\omega}= \prod_{n=1}^{\infty} \cos \frac{\omega}{2^{n}}$ (3)

... to obtain from (1) and (2)...

$\displaystyle \Phi(\omega)= e^{-i\ \frac{\omega}{2}}\ \frac{\sin \omega}{\omega}$ (4)

... so that Y is uniformly distributed between 0 and 1...

Kind regards

$\chi$ $\sigma$

 

[bennyzadir]

Thank you for your answer. Do you have any idea for part b) ?

 

[chisigma]

If $p \ne \frac{1}{2}$ the task becomes a little more complex. In that case You have...

$\displaystyle \varphi_{n}(x)= p\ \delta(x) + (1-p)\ \delta (x-\frac{1}{2^{n}}) \implies \Phi_{n} (\omega)= (1-p)\ e^{- i \frac{\omega}{2^{n}}}\ (1+ \frac{p}{1-p}\ e^{i \frac{\omega}{2^{n}}})$ (1)

... and now You have to valuate the 'infinite product'...

$\displaystyle \Phi(\omega)= \prod_{n=1}^{\infty} \Phi_{n}(\omega)$ (2)

What You can demonstrate is that the infinite product (2) converges because converges the term...

$\displaystyle \prod_{n=1}^{\infty} (1+ \frac{p}{1-p}\ e^{i \frac{\omega}{2^{n}}})$ (3)

... and that is true because converges the series...

$\displaystyle \sum_{n=1}^{\infty} e^{i \frac{\omega}{2^{n}}}$ (4)

The effective computation of (2) is a different task that requires a little of efforts...

Kind regards

$\chi$ $\sigma$

 

[blue_raver22]

a) To prove that Y is uniformly distributed on the interval [0,1], we need to show that for any interval [a,b] with 0≤a<b≤1, the probability that Y falls within this interval is equal to the length of the interval, i.e. b-a.

First, note that Y can only take on values in the interval [0,1] since it is a sum of terms that are either 0 or 2^-n. Therefore, we can write the probability that Y falls within [a,b] as:

P(a≤Y≤b) = P(Y≤b) - P(Y≤a)

= P(Y≤b) - P(Y<a)

= P(Y≤b) - P(Y=0)

= P(Y≤b) - (1-p)^∞

= P(Y≤b) - 0 (since p=1/2)

= P(Y≤b)

Now, we can use the fact that Y is a sum of independent random variables to calculate this probability. Since each X_i is either 0 or 1 with equal probability, the probability that Y takes on a particular value y is equal to the number of ways we can get that value divided by the total number of possible outcomes.

Therefore, P(Y=y) = 1/2^k where y is a sum of exactly k terms. Since the values of Y are unique, we can rewrite P(Y≤b) as:

P(Y≤b) = Σ P(Y=y) = Σ 1/2^k

where the summation is taken over all y such that y≤b. This is a geometric series with a common ratio of 1/2 and a starting term of 1. Therefore, we can use the formula for the sum of a geometric series to get:

P(Y≤b) = 1/1-1/2 = 2/2-1 = 1

Similarly, we can calculate P(Y≤a) and get the same result of 1. Therefore, we have shown that P(a≤Y≤b) = P(Y≤b) - P(Y≤a) = 1-1 = 0. This is true for any interval [a,b] with 0≤a<b≤1, which means that