# Probabilty of Elements being smaller than a specific value

Right now i am looking at the following statement, but i cant grasp why it is correct. Can somebody help?
"If we look at F0 uniformly distributed (and, say, pairwise independent) elements of [0, 1], we expect about t of them to be smaller than t/F0."

Kind Regards, Ilian

Consider defining indicator random variables $$X_1, X_2, ..., X_{F_0}$$ such that

$$X_i = \begin{cases} 1 \text{ if } x_i < \frac{t}{F_0} \\ 0 \text{ otherwise } \end{cases}$$

Then, we can count the number of elements $$< \frac{t}{F_0}$$ as follows

$$[ \text{# of Elements} < \frac{t}{F_0} ]= \sum_{i=1}^{F_0} X_i$$

Thus,

$$\mathbb{E} [\text{# of Elements} < \frac{t}{F_0}] = \mathbb{E} [\sum_{i=1}^{F_0} X_i] \\ = \sum_{i=1}^{F_0} \mathbb{E}[X_i]$$

By linearity of expectation. Note that by virtue of $$X_i$$'s being indicator random variables, and independently identically distributed, we have that

$$\mathbb{E}[X_i] = 1 \cdot \mathbb{P}[X_i=1] + 0 \cdot \mathbb{P}[X_i=0] = \frac{t}{F_0}$$

For all $$1 \leq i \leq F_0$$. Substituing, we get that

$$\mathbb{E} [\text{# of Elements} < \frac{t}{F_0}] = \sum_{i=1}^{F_0} \mathbb{E}[X_i] = \sum_{i=1}^{F_0} \frac{t}{F_0} = t$$

• Thank you very much! Nov 16, 2022 at 15:22

If you cut the interval in 5 pieces of length 1/5 and pick one then you would expect an element drawn uniformly at random to have chance 1/5 to belong to it. So for, say t =3, an interval of length 3/N the chance is 3/N. Hence you would expect 3 out of N draws to belong to it. Apply this to the interval with boundary 0 and 3/N.

• Also to you: Thank you very much! Nov 16, 2022 at 15:22