Jonobro said:Homework Statement
Let X∼Geometric(1/3), and let Y=|X−5|. Find the range and PMF of Y.
Homework Equations
Px(k) = p(1-p)^(k-1) for x=1,2,3...
3. My attempt at a solution
I set up the PMF for Px
Px(k) = 1/3(2/3)^(k-1) for k = 1,2,3,...
However I don't know how to convert this to Y
The range of Y for X ∼ Geometric(1/3) is all positive integers from 1 to infinity.
The PMF of Y for X ∼ Geometric(1/3) is calculated using the formula P(Y = k) = (1-p)^k * p, where p is the probability of success (1/3 in this case) and k is the number of trials.
A PMF graph for X ∼ Geometric(1/3) will have a peak at 1 and then gradually decrease as the value of k increases, forming a right-skewed distribution.
The expected value of Y for X ∼ Geometric(1/3) is equal to 1/p, so in this case it is 3.
No, the range of Y for X ∼ Geometric(1/3) cannot be negative as it is a discrete probability distribution only defined for positive integers.