Question

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

Answer 1

l=k-1
p(y) =1/k

Answer 2

c=1
fz(z)=1

Answer 3

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.

Answer 4

l = k-1 Violates the condition py(y) > 0 for y = l +1

Answer 5

l = [k]

Answer 6

l = k
py(y) = 1/(k+1)

Answer 7

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

Answer 8

Actually, forget it, 2. is 1/(k+1) as it includes 0