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).