# Thread: sum of Bernoulli trials

1. ## sum of Bernoulli trials

Hoi, i dont see why the following is true

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

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

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

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

I'm a bit bewildered...:/

2. ## 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).

3. ## Re: sum of Bernoulli trials

I think i got it now...

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

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