Let's consider the probability distribution function (PDF) of X. Since X is uniformly distributed on [0,k], the PDF of X is given by:
fX(x) = 1/k, for 0 ≤ x ≤ k
fX(x) = 0, otherwise
Given that Y = ⌊X⌋, we want to find the probability mass function (PMF) of Y, denoted pY(y).
We can start by considering the possible values of Y. Since Y represents the largest integer not exceeding X, the possible values of Y are y = 0, 1, 2, ..., ⌊k⌋. Let's denote ⌊k⌋ as ℓ.
Now, we need to find the probability pY(y) for y = 0, 1, 2, ..., ℓ.
For y = 0:
pY(0) = P(Y = 0)
= P(0 ≤ X < 1)
= ∫[0,1) fX(x) dx
= ∫[0,1) 1/k dx
= 1/k * (x)|[0,1)
= 1/k
For y = 1:
pY(1) = P(Y = 1)
= P(1 ≤ X < 2)
= ∫[1,2) fX(x) dx
= ∫[1,2) 1/k dx
= 1/k * (x)|[1,2)
= 1/k
We can see that for y = 0 and y = 1, pY(y) = 1/k.
For y = 2:
pY(2) = P(Y = 2)
= P(2 ≤ X < 3)
= ∫[2,3) fX(x) dx
= ∫[2,3) 1/k dx
= 1/k * (x)|[2,3)
= 1/k
We can see that for y = 2, pY(y) = 1/k.
By observing this pattern, we can conclude that for y = 0, 1, 2, ..., ℓ, pY(y) = 1/k.
Therefore, the value of ℓ is ⌊k⌋, and the PMF of Y for y = 0, 1, 2, ..., ℓ is given by:
pY(y) = 1/k, for y = 0, 1, 2, ..., ⌊k⌋
Let k be a positive integer and let X be a continuous random variable that is uniformly distributed on [0,k]. For any number x, denote by ⌊x⌋ the largest integer not exceeding x. Similarly, denote frac(x)=x−⌊x⌋ to be the fractional part of x. The following are two properties of ⌊x⌋ and frac(x):
x = ⌊x⌋+frac(x)
⌊x⌋ ≤ x<⌊x⌋+1,
frac(x) ∈ [0,1).
For example, if x=2.91, then ⌊x⌋=2 and frac(x)=0.91.
Let Y=⌊x⌋ and let pY(y) be its PMF. There exists some nonnegative integer ℓ such that pY(y)>0 for every y∈{0,1,…,ℓ}, and pY(y)=0 for y≥ℓ+1. Find ℓ and pY(y) for y∈{0,1,…,ℓ}. Your answer should be a function of k.
1 answer