Probability related to cumulative distribution function

[songoku]

Homework Statement

Please see below

Relevant Equations

F(X) = P(X ≤ x)

I have tried to answer all the questions but I am not that sure with my answer.

That's the graph of $F_X (x)$ (I think)

(i) P (X ≤ i) = $\frac{i^2}{N^2}$ and P(X < i) = 0
All of these are based on the graph

(ii) P(X = i) = P(X ≤ i) - P(X < i) = $\frac{i^2}{N^2}$

Are my answers correct? Thanks

 

[FactChecker]

Homework Statement: Please see below
Relevant Equations: F(X) = P(X ≤ x)

View attachment 333620

I have tried to answer all the questions but I am not that sure with my answer.

View attachment 333621
That's the graph of $F_X (x)$ (I think)

(i) P (X ≤ i) = $\frac{i^2}{N^2}$ and P(X < i) = 0
All of these are based on the graph

If there are several integers between 0 and i, they have positive probability values. so P(X<i) > 0.

(ii) P(X = i) = P(X ≤ i) - P(X < i) = $\frac{i^2}{N^2}$

Are my answers correct? Thanks

No. Notice that your diagram only has one i < N, but there might be several others. Also, the sum of the probabilities must equal 1, so your diagram is missing a lot of probability.

 

[songoku]

If there are several integers between 0 and i, they have positive probability values. so P(X<i) > 0.

No. Notice that your diagram only has one i < N, but there might be several others. Also, the sum of the probabilities must equal 1, so your diagram is missing a lot of probability.

Ah, I see. Now I understand the question

Revised attempt:
(i)
$$P (X ≤ i) = \frac{i^2}{N^2}$$

$$P(X < i) =
\begin{cases}
0 & \text{if } i= 0 \\
\frac{(i-1)^2}{N^2} & \text{if } i>0
\end{cases}
$$

(ii)
$$P(X = i) =
\begin{cases}
0 & \text{if } i= 0 \\
\frac{2i-1}{N^2} & \text{if } i>0
\end{cases}
$$

For P(X < i) and P(X = i), is there an answer not involving piecewise function? Thanks

 

[anuttarasammyak]

just resuming your effort : using Heaviside step function H
F_X(x)=\frac{1}{N^2}\sum_{i=1}^N (i^2-(i-1)^2)H_1(x-i)
Probability density is by differentiation
p(x)=\frac{1}{N^2}\sum_{i=1}^N (i^2-(i-1)^2)\delta(x-i)
Probability for digits are by integrating p(x) around x=i
P(i)=\frac{i^2-(i-1)^2}{N^2}=\frac{2i-1}{N^2}
for $1 \leq i \leq N$. Otherwise p(x)=0, P(i)=0.

 

[songoku]

Thank you very much for the help and explanation FactChecker and anuttarasammyak