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.
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⌋
To find ℓ and pY(y) for y ∈ {0, 1, ..., ℓ}, we need to understand the distribution of Y, which is ⌊x⌋.
Since X is a continuous random variable that is uniformly distributed on [0, k], the distribution of Y will depend on the value of k.
Let's analyze the possible values of ⌊x⌋ based on the interval [0, k].
1. When x is in the interval [0, 1):
In this case, ⌊x⌋ will always be 0, as it is the largest integer not exceeding x. Therefore, pY(0) = 1, and pY(y) = 0 for y > 0.
2. When x is in the interval [1, 2):
Here, ⌊x⌋ will always be 1, and pY(1) = 1. pY(y) = 0 for y > 1.
3. Continuing this pattern, for x in the interval [n, n+1), ⌊x⌋ will always be n, and pY(n) = 1. pY(y) = 0 for y > n.
To determine the value of ℓ, we need to find the largest integer n such that n ≤ k. Therefore, ℓ = ⌊k⌋.
Now that we know the value of ℓ, we can write the probability mass function (PMF) pY(y) for y ∈ {0,1,...,ℓ} as follows:
- For y = 0, 1, ..., ℓ, pY(y) = 1 for y ≤ ℓ, and pY(y) = 0 for y > ℓ.
Therefore, the value of ℓ is ⌊k⌋, and pY(y) for y ∈ {0,1,...,ℓ} is given by:
pY(y) = {
1 if 0 ≤ y ≤ ⌊k⌋,
0 otherwise.
}
So, ℓ = ⌊k⌋ and pY(y) = 1 if 0 ≤ y ≤ ⌊k⌋, and pY(y) = 0 otherwise.