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):
\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
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)
To find the value of 𝓁 and 𝓅𝒀(𝒚) for 𝒚 ∈ {0,1, ..., 𝓁}, we need to consider the probability of 𝑌 = 𝒚, where 𝑌 = ⌊𝑥⌋.
Since 𝑋 is uniformly distributed on [0, 𝑘], the probability density function (PDF) of 𝑋 is given by:
𝑓𝑋(𝑥) = 1/𝑘 if 0 ≤ 𝑥 ≤ 𝑘
= 0 otherwise
To find 𝑝𝒀(𝒚), we need to integrate the PDF of 𝑋 over the interval [𝑦, 𝑦 + 1]:
𝑝𝒀(𝒚) = ∫[𝑦,𝑦+1] 𝑓𝑋(𝑥) 𝑑𝑥
Since 𝑋 is uniformly distributed, 𝑓𝑋(𝑥) = 1/𝑘 for 𝑥 in [0, 𝑘]. We can split the integral into two parts:
𝑝𝒀(𝒚) = ∫[𝑦,𝑦+1] 1/𝑘 𝑑𝑥
Since ∫ 𝑑𝑥 = 𝑥, the integral becomes:
𝑝𝒀(𝒚) = (1/𝑘) * (𝑥∣[𝑦,𝑦+1])
𝑝𝒀(𝒚) = (1/𝑘) * (𝑥∣[𝑦,𝑦+1]) = 1/𝑘
Since there are 𝑘 possible integer values for 𝒚, 𝓁 = 𝑘 - 1, and 𝓅𝒀(𝒚) = 1/𝑘 for 𝒚 ∈ {0,1, ..., 𝑘-1}.
Thus, 𝓁 = 𝑘 - 1 and 𝓅𝒀(𝒚) = 1/𝑘 for 𝒚 ∈ {0,1, ..., 𝑘-1}.
Now, to find the value of 𝑐 and 𝑓𝒁(𝑧) for 𝑧 in (0,𝑐), we need to consider the probability density function (PDF) of 𝑍 = 𝐟𝑟𝑎𝑐(𝑥).
Since 𝑍 = 𝐟𝑟𝑎𝑐(𝑋), where 𝑋 is uniformly distributed on [0, 𝑘], the distribution of 𝑍 is also uniform on [0, 1).
Therefore, 𝑓𝑍(𝑧) = 1 for 𝑧 in (0, 1), and 𝑐 = 1.
So, 𝑐 = 1, and 𝑓𝒁(𝑧) = 1 for 𝑧 in (0, 1).