 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2023/discrete-mathematics-probability-and-continuous-mathematics/9/b/iii/solution

For

k \geq 1

, the event

{⌊ X ⌋ \geq k}

equals

{X \geq k}

P (⌊ X ⌋ \geq k) = k^{1 - t}, P (⌊ X ⌋ = k) = k^{1 - t} - (k + 1)^{1 - t} (k \geq 1) .

The mass formula follows by subtracting consecutive tail probabilities, or by integrating over

[k, k + 1)

. With the paper's convention

0 \in N

, include the separate endpoint values $P (⌊ X ⌋ \geq 0) = 1$ and $P (⌊ X ⌋ = 0) = 0$ , since

X \geq 1

almost surely. Substituting

k = 0

into the power expression would be invalid.

 New to topics? Read the docs here!