# sum of Bernoulli trials

• Oct 15th 2012, 03:46 AM
Dinkydoe
sum of Bernoulli trials
Hoi, i dont see why the following is true

Let $\mathbb{P}(X_k =1)=\mathbb{P}(X_k=0)=1/2$ ( i.i.d. Bernoulli trials)

and consider $Y = 3\sum_{k=1}^{\infty}4^{-k}X_k$ ( Then clearly $0\leq Y \leq 1$)

Apparantly the distribution function F of Y is constant on the interval $(1/4,3/4)$ and satisfies
$F(x) = 1-F(x)$ and

for $x< 1/4$ it satisfies $F(x)=2F(x/4)$

I'm a bit bewildered...:/
• Oct 15th 2012, 06:36 PM
chiro
Re: sum of Bernoulli trials
For the first one, can you show that the median occurs at that point? If your distribution is symmetric, can you show the mean occurs at that point? (Hint: your distribution has 1/2 for a 0 value and 1/2 for a 1 value for every X_k).
• Oct 16th 2012, 12:55 AM
Dinkydoe
Re: sum of Bernoulli trials
I think i got it now...

$P(X \leq 1/4) = P(X_1=0) = 1/2$ and $P(X\leq 3/4 ) = 1- \mathbb{P}(X\geq 3/4) = 1 - \mathbb{P}(X_1=1) = 1/2$

I think i see the other 2 as well...