# Question "The value of the CDF F(x), where x is a particular outcome"

The value of the CDF F(x), where x is a particular outcome, for a discrete uniform distribution:

Answer is lies b/w 0 and 1

My question is why is the answer “sums to 1” incorrect. Is it because the sum probability function P(x) is 1 and not F(x).
Correct me if i am wrong

Example: discrete uniform for 1, 2, 3, 4, and 5

F(0)=0
F(1) = 1/5
F(2) = 2/5

F(5) = 1.0

The CDF gives the sum of the probabilities for all X<=a. The only way to get F(x)=1 is for X>=a.

So in both

1. x<= a
and
2. x >=a
We have a scenario where x=a and therefore F(x) = 1.
So is the Sum of CDF = 1 or not

You are confusing F(x) with lim x->inf F(x). They are not the same.

By definition, F(x) = Pr(X <= x), where X is a random variable.
That is, F(x) is the probability that X turns out to have a value of x or below.
It only reaches the value 1 if you sum over all possible outcomes, which may or may not be infinite (in the case of continuous random variables, you’d have to take an integral).

For example, consider the roll of a fair, 6-sided die. Each side has an equal probability of being landed, 1/6.
Pr(X = 1) = Pr(X = 2) = Pr(X = 3) = Pr(X = 4) = Pr(X = 5) = Pr(X = 6) = 1/6
F(1) = Pr(X = 1) = 1/6
F(2) = Pr(X = 1) + Pr(X = 2) = 2/6
only when we sum over all possible outcomes that F(x) becomes 1:
F(6) = Pr(X = 1) + Pr(X = 2) + … + Pr(X= 6) = 1/6 + … + 1/6 = 6/6 = 1

The upper limit for F(x) is thus 1, or 100%.
The lower limit for F(x) is 0, and happens when an outcome cannot happen. For example, in the die example, F(0) = 0, since X can’t be 0 no matter how many times you roll the die - it can only take the values 1, 2, 3, 4, 5, or 6.

edit:
As for the option that says F(x) sums to 1: i think you misunderstood the statement.
It meant to say F(x) = 1 for some unspecified outcome x. This is not necessarily true, as was shown in previous examples above.