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.
1. 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=
pY(y) =
2. 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 =
fZ(z) =
1. To find ℓ, we need to consider the largest value of ⌊x⌋ when x is in the interval [0, k]. Since ⌊x⌋ is the largest integer not exceeding x, when x = k, we have ⌊x⌋ = k - 1. Therefore, ℓ = k - 1.
Now let's find pY(y) for y ∈ {0,1,...,ℓ}. Since Y = ⌊x⌋, for any value of y in the range [0, ℓ], pY(y) represents the probability that ⌊x⌋ takes on the value of y.
Since X is uniformly distributed on [0, k], the probability density function (PDF) of X is fX(x) = 1/k for 0 ≤ x ≤ k, and 0 otherwise.
To find pY(y), we need to integrate fX(x) over the range of x values for which ⌊x⌋ = y. This occurs when y ≤ x < y + 1.
∫[y,y+1) fX(x) dx = ∫[y,y+1) (1/k) dx = (x/k)∣[y,y+1) = (y+1 - y)/k = 1/k
Therefore, pY(y) = 1/k for y ∈ {0,1,...,ℓ}.
2. To find c, we need to consider the range of values that the fractional part frac(x) can take. Since frac(x) = x - ⌊x⌋, and ⌊x⌋ takes on integer values in the range [0,ℓ], frac(x) takes on values in the range [0,1).
Therefore, c = 1.
Now let's find fZ(z) for z ∈ (0,c). Since Z = frac(x), we need to find the probability density function (PDF) of Z.
To find fZ(z), we can differentiate the cumulative distribution function (CDF) of Z:
FZ(z) = P(Z ≤ z) = P(frac(x) ≤ z) = P(0 ≤ x - ⌊x⌋ ≤ z)
Since x is uniformly distributed on [0,k], the probability that the difference between x and its floor value ⌊x⌋ is less than or equal to z is equal to the length of the interval [0,k] for which the difference is less than or equal to z.
To find this length, we consider two cases:
Case 1: z < 1 - ⌊z⌋ (z is smaller than the distance from ⌊z⌋ to 1)
In this case, the length of the interval [0,k] for which the difference between x and ⌊x⌋ is less than or equal to z is equal to k.
Case 2: z ≥ 1 - ⌊z⌋ (z is greater than or equal to the distance from ⌊z⌋ to 1)
In this case, the length of the interval [0,k] for which the difference between x and ⌊x⌋ is less than or equal to z is equal to k - (1 - ⌊z⌋).
Combining these two cases, we can write the CDF as:
FZ(z) = k if z < 1 - ⌊z⌋
FZ(z) = k - (1 - ⌊z⌋) if z ≥ 1 - ⌊z⌋
To find the PDF, we differentiate the CDF with respect to z:
fZ(z) = d/dz FZ(z)
Since the CDF is constant within certain intervals, the derivative will be zero in those intervals:
fZ(z) = 0 if z < 1 - ⌊z⌋
fZ(z) = 1 if z = 1 - ⌊z⌋
fZ(z) = 0 if z > 1 - ⌊z⌋
Therefore, c = 1 and fZ(z) = 1 for z ∈ (0,c).
1. To find ℓ and pY(y), we need to consider the possible values of ⌊x⌋ for the given range of X = [0, k].
Since X is uniformly distributed on [0, k], the values of X are equally likely within this interval.
For any given value of X, ⌊x⌋ will be the largest integer less than or equal to X.
So, ⌊x⌋ will take values from 0 to k-1, inclusive. Therefore, ℓ = k-1.
The probability distribution function pY(y) is the probability that ⌊x⌋ takes on the value y.
Since ⌊x⌋ is the largest integer less than or equal to X, the probability that ⌊x⌋ is equal to y is the probability that X lies in the interval [y, y+1).
Since X is uniformly distributed on [0, k], this probability is (y+1 - y)/k = 1/k.
For y∈{0,1,…,ℓ}, pY(y) = 1/k.
Therefore,
ℓ = k - 1
pY(y) = 1/k for y∈{0,1,…,ℓ}
2. To find c and fZ(z), we consider the PDF of Z = frac(x).
The fractional part of x, frac(x), lies in the interval [0,1).
Since X is uniformly distributed on [0, k], the values of frac(x) will also be uniformly distributed on [0,1).
The PDF, fZ(z), represents the probability density of frac(x) taking on a particular value z.
Since frac(x) is uniformly distributed on [0,1), the PDF fZ(z) will be constant within this interval.
Therefore, fZ(z) is a constant value for z∈(0,1), and it is equal to 1.
So, c = 1 and fZ(z) = 1 for z∈(0,1).
To find the value of ℓ and pY(y) for Y=⌊x⌋, we need to understand the distribution of Y. Since X is uniformly distributed on [0,k], the range of possible values for Y is {0,1,2,...,⌊k⌋}.
Property 2 tells us that ⌊x⌋ is always less than or equal to x and less than ⌊x⌋ + 1. This means that Y = ⌊x⌋ will take on integer values from 0 to ⌊k⌋.
However, since Y can take on the value ⌊k⌋ only if X is exactly equal to k, the probability of Y being ⌊k⌋ will be 0 if k is not an integer. Therefore, ℓ = ⌊k⌋ - 1.
To find pY(y) for y∈{0,1,…,ℓ}, we need to consider the probability of Y being each value y. Since X is uniformly distributed, the probability of Y being y is the same as the probability of X being in the interval [y, y+1).
Since X is uniformly distributed on [0,k], the length of the interval [y, y+1) is 1. Therefore, pY(y) = 1/k for y∈{0,1,…,⌊k⌋-1}.
To summarize:
ℓ = ⌊k⌋ - 1
pY(y) = 1/k for y∈{0,1,…,⌊k⌋-1}
Now let's move on to Z=frac(x) and find the value of c and fZ(z).
Property 2 tells us that frac(x) is always between 0 and 1. Therefore, the range of possible values for Z is (0,1).
To find c, we need to determine the maximum value of fZ(z) in the range (0,1). Since Z is the fractional part of X, it takes on all possible values between 0 and 1 uniformly.
Therefore, the maximum value of fZ(z) occurs at any point within the range (0,1), and c can be any value between 0 and 1.
To summarize:
c can be any value between 0 and 1
fZ(z) is a constant distribution between 0 and c, which means fZ(z) = 1/c for z∈(0,c).
To get the precise values of ℓ, pY(y), c, and fZ(z), you will need to provide a specific value for k.