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.
ℓ=
unanswered
pY(y)=
unanswered
Let Z=frac(x) and let fZ(z) be its PDF. There exists a real number c such that fZ(z)>0 for every z∈(0,c), and fZ(z)=0 for every z>c. Find c, and fZ(z) for z∈(0,c).
c=
unanswered
fZ(z)=
unanswered
c=1
fz(z)=1
l=k-1
p(y) =1/k
l=k-1?
How?
Suppose we have k=2 so that y can now be 0,1, and 2. So l should be 2 in this case.
P(y)=0 for l+1. So if we have l=k-1, we would have l=1 when k=2. This would violate the condition we have set i.e P(y)=0 for l+1 since y can clearly be 2 if k=2.
l = [k]
l = k-1 Violates the condition py(y) > 0 for y = l +1
Actually, forget it, 2. is 1/(k+1) as it includes 0
l = k
py(y) = 1/(k+1)
1. k
2. 1/k (remember it's a uniform distribution and it has to be a valid PMF)
3. c
4. 1
It seems quite easy to be true, so I'm not sure about this
To find the values of ℓ and pY(y) for y∈{0,1,…,ℓ}, we need to understand the distribution of Y=⌊x⌋.
In this case, since X is uniformly distributed on [0,k], the possible values of ⌊x⌋ are {0,1,2,...,k-1}. The largest possible value for ⌊x⌋ is k-1 because the interval [0,k] does not include k itself. Therefore, the value of ℓ is k-1.
Now, let's find pY(y) for y∈{0,1,…,ℓ}. Since ⌊x⌋ is the largest integer not exceeding x, it can take the values {0,1,2,...,k-1} with equal probability.
Since ⌊x⌋ is an integer, pY(y) represents the probability that an integer value ⌊x⌋ is equal to y. Since ⌊x⌋ takes values from 0 to k-1 with equal probability, the probability can be calculated as 1/(k-1+1) = 1/k.
Therefore, ℓ = k-1, and pY(y) = 1/k for y∈{0,1,…,ℓ}.
Now, let's move on to finding the value of c and fZ(z) for z∈(0,c).
Z=frac(x) represents the fractional part of x, which ranges from 0 to 1.
Since Z takes values from 0 to 1, the PDF fZ(z) will have the same value across this interval. To find c, we need to determine the maximum value of z for which fZ(z) is non-zero.
In this case, since Z is continuous and uniformly distributed on [0,1), the PDF fZ(z) is a constant over the entire interval [0,1). Therefore, c = 1.
And since fZ(z) is constant over the interval (0,1), fZ(z) = 1 for z∈(0,1).
Therefore, c = 1, and fZ(z) = 1 for z∈(0,1).