Let $X_1,X_2,\dots$be a sequence of independent random variables having the probability mass function$P\left\{X_n=0\right\}=P\left\{X_n=2\right\}=1/2,n\ge 1$

The random variable $X=\sum _{n=1}^{\infty }{X}_n/3^n$is said to have the Cantor distribution.

Find $E\left[X\right]$and$Var\left(X\right).$

Sequence of an independent random variable is $X_1,X_2,\dots$

The probability mass function is $P\left\{X_n=0\right\}=P\left\{X_n=2\right\}=1/2,n\ge 1$

Cantor distribution's random variable$X=\sum _{n=1}^{\infty }{X}_n/3^n$

From the given definition, one has that:

$X=\sum _{n=1}^{\infty }\frac{X_n}{3^n}$

$E\left[X\right]=\sum _{n=1}^{\infty }E\left[\frac{X_n}{3^n}\right]$

$E\left[X\right]=\sum _{n=1}^{\infty }\frac{1}{3^n}E\left[X_n\right]$

Each of the random variables, $X_1,X_2\dots$is equally likely to be either 0 or $\frac{1}{2}.$.

Hence, the expectation value of each of the independent variables is given by:

$E\left[X_n\right]=\frac{1}{2}2+0\frac{1}{2}$

$E\left[X_n\right]=1$

Using the above result, one has that:

$E\left[X\right]=\sum _{n=1}^{\infty }\frac{1}{3^n}$

$=\frac{1}{2}$

Therefore, the mean is$E\left[X\right]=\frac{1}{2}$

Compute, variance one has that:

$X=\sum _{n=1}^{\infty }\frac{X_n}{3^n}$

$Var\left(X\right)=\sum _{n=1}^{\infty }Var\left(\frac{X_n}{3^n}\right)$

$Var\left(X\right)=\sum _{n=1}^{\infty }\frac{1}{3^{2n}}Var\left(X_n\right)$

Now computing the summation term, one has that:

$Var\left(X_n\right)=E\left[X_n^2\right]-E{\left[X_n\right]}^2$

$Var\left(X_n\right)=\left(0·\frac{1}{2}+\frac{1}{2}·4\right)-1$

$Var\left(X_n\right)=1$

Using the above result, one has that:

$\mathrm{Var}\left(X\right)=\sum _{n=1}^{\mathrm{\infty }} \frac{1}{3^{2n}}$

$=\frac{1}{8}$

Hence, the mean value is $E\left[X\right]=\frac{1}{2}$.

And the variance is $Var\left(X\right)=\frac{1}{8}.$