# Thread: P(y=1), e[y]

2. ## Re: P(y=1), e[y]

For any probability distribution we must have total probability equal to 1 Here, that means that $\displaystyle \sum_{i= 0}^\infty \frac{\alpha}{2^i}= \alpha\sum_{i=0}^\infty \left(\frac{1}{2}\right)^i= 1$. That sum is a geometric series, $\displaystyle \sum_{i= 0}^\infty r^i$ with r= 1/2. It's sum is $\displaystyle \frac{1}{1- r}= \frac{1}{1- \frac{1}{2}}= \frac{1}{\frac{1}{2}}= 2$. So we must have $\displaystyle 2\alpha= 1$.

If Y= X (mod 3)= 1 then X is 1, 4, 7, 10, etc so that $\displaystyle P(1)= \frac{\alpha}{2}+ \frac{\alpha}{2^4}+ \frac{\alpha}{2^7}+ \frac{\alpha}{2^{10}}+ \cdot\cdot\cdot= $$\displaystyle \frac{\alpha}{2}\left(1+ \frac{1}{2^3}+ \frac{1}{2^6}+ \frac{1}{2^9}+ \cdot\cdot\cdot\right)=$$\displaystyle \frac{\alpha}{2}\left(1+ \frac{1}{2^3}+ \frac{1}{(2^3)^2}+ \frac{1}{(2^3)^3}+ \cdot\cdot\cdot\right)$. That sum is again a geometric series with $\displaystyle r= \frac{1}{2^3}= \frac{1}{8}$.

3. ## Re: P(y=1), e[y]

Hi, thanks. But can you explain a bit? i do not full understand how i can get probability P[X],
for geometric distribution
1. if i want get E[X]=1/p, i have to first get probability P[X]

2. I think P(Y=1)=a/2 = 1/4 (2a=1, a=1/2)

E[Y] = 1/p, but i need get p[Y]

4. ## Re: P(y=1), e[y]

A "geometric distribution" is, by definition, a distribution of the form $\displaystyle P(X= k)= (1- p)^{k-1}p$. If you are still talking about the distribution in your first post, $\displaystyle P(X= I)= \frac{\alpha}{2^i}= \alpha \left(\frac{1}{2}\right)^i=$ then that is a geometric distribution with p= 1/2 so that 1- p= 1/2 also: $\displaystyle P(X= k)= (1/2)^{k-1}(1/2)= (1/2)^k$.

5. ## Re: P(y=1), e[y]

hi, i study your advice again, here is my understand:
first problem: a = 1/2
p= 1/2
E[X] = (1-p)/p = 1 (k∈{0,1,2,3,...})

second problem: Y = X mod 3
r = 1/8
a/1-r = 1
a = 7/8
P(Y=1) = a/2 = 7/16
E[Y] = (1-p)/p = 9/7

do you think my understand correct?