# Sum of random variables

##### New member
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 continous but not absolutely continous 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!

#### chisigma

##### Well-known member
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 continous but not absolutely continous 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!
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

$\chi$ $\sigma$

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

#### chisigma

##### Well-known member
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

$\chi$ $\sigma$