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 \lfloor x\rfloor the largest integer not exceeding x. Similarly, denote \text {frac}(x)=x-\lfloor x\rfloor to be the fractional part of x. The following are two properties of \lfloor x\rfloor and \text {frac}(x):

Question

\displaystyle \displaystyle x \displaystyle = \displaystyle \lfloor x\rfloor +\text {frac}(x)
\displaystyle \lfloor x\rfloor \displaystyle \leq \displaystyle x<\lfloor x\rfloor +1,
\displaystyle \text {frac}(x) \displaystyle \in \displaystyle [0,1).
For example, if x=2.91, then \lfloor x\rfloor =2 and \text {frac}(x)=0.91.

Let Y=\lfloor x\rfloor and let p_ Y(y) be its PMF. There exists some nonnegative integer \ell such that p_ Y(y)>0 for every y\in \{ 0,1,\dots ,\ell \}, and p_ Y(y)=0 for y\geq \ell +1. Find \ell and p_ Y(y) for y\in \{ 0,1,\dots ,\ell \}. Your answer should be a function of k.

\ell =\quad
unanswered

p_ Y(y)=\quad
unanswered

Let Z=\text {frac}(x) and let f_ Z(z) be its PDF. There exists a real number c such that f_ Z(z)>0 for every z\in (0,c),\, and f_ Z(z)=0 for every z> c. Find c,\, and f_ Z(z) for z\in (0,c).

c=\quad
unanswered

f_ Z(z)=\quad

Answer 1

First, let's find the probability mass function (PMF) of Y.

Since X is uniformly distributed on [0, k], the probability that X takes on a value in the interval [n, n+1) is equal to the length of that interval divided by the length of the entire interval [0, k].

The length of the interval [n, n+1) is 1, and the length of [0, k] is k, so the probability that X takes on a value in [n, n+1) is 1/k.

Therefore, p_Y(y), the probability that Y is equal to y, is equal to the probability that X takes on a value in the interval [y, y+1), which is 1/k.

Since p_Y(y) = 0 for y ≥ ℓ + 1, we want to find the largest integer ℓ such that p_Y(y) ≠ 0 for y = 0, 1, ..., ℓ.

For y = 0, 1, ..., ℓ, p_Y(y) = 1/k ≠ 0.

Therefore, ℓ = k - 1.

So we have ℓ = k - 1 and p_Y(y) = 1/k for y = 0, 1, ..., k - 1.

Next, let's find the probability density function (PDF) of Z.

Since Z = frac(X), Z has a uniform distribution on [0, 1).

The PDF of a uniform distribution on [0, 1) is constant over its support (in this case, [0, 1)) and is equal to 1/c, where c is the length of the support.

Since the support of Z is [0, 1), c = 1.

Therefore, c = 1 and f_Z(z) = 1 for z ∈ (0, 1).

So we have c = 1 and f_Z(z) = 1 for z ∈ (0, 1).

In summary, the values we found are:

ℓ = k - 1

p_Y(y) = 1/k for y = 0, 1, ..., k - 1

c = 1

f_Z(z) = 1 for z ∈ (0, 1)