variance of discrete random variables

Question 1

spread

Answer 1

The expected value (mean) of a random variable is a measure of location or central tendency.
If you had to summarize a random variable with a single number, the mean would be a good
choice. Still, the mean leaves out a good deal of information.

Question 2

Give an example of random variables X and Y that have mean 0 but have a very different PMS spread

Answer 2

Question 3

Variance

Answer 3

Taking the mean as the center of a random variable’s probability distribution, the variance
is a measure of how much the probability mass is spread out around this center.

Question 4

Formal definition of variance

Answer 4

Question 5

The standard deviation 𝜎 of 𝑋 is defined by

Answer 5

Question 6

note on the units of values

Answer 6

1. 𝜎 has the same units as 𝑋.
2. Var(𝑋) has the same units as the square of 𝑋. So if 𝑋 is in meters, then Var(𝑋) is in meters squared.

Because 𝜎 and 𝑋 have the same units, the standard deviation is a natural measure of spread.

Question 7

Answer 7

Question 8

Answer 8

Question 9

The variance of a Bernoulli(𝑝) random variable

Answer 9

Bernoulli random variables are fundamental, so we should know their variance.

If 𝑋 ∼ Bernoulli(𝑝) then
Var(𝑋) = 𝑝(1 − 𝑝)

Question 10

Prove the variance of Bernoulli(p)

Answer 10

Question 11

independence of discrete random variables

Answer 11

Question 12

properties of variance

Answer 12

For Property 1, note carefully the requirement that 𝑋 and 𝑌 are independent. We will
return to the proof of Property 1 in a later class.

Property 3 gives a formula for Var(𝑋) that is often easier to use in hand calculations. The
computer is happy to use the definition! We’ll prove Properties 2 and 3 after some examples.

Question 13

Answer 13

Question 14

Use Property 3 to compute the variance of 𝑋 ∼ Bernoulli(𝑝).

Answer 14

Question 15

proof of property 2

Answer 15

Question 16

proof of property 3

Answer 16

Question 17

variance of binomial(n, p)

Answer 17

Suppose 𝑋 ∼ binomial(𝑛, 𝑝). Since 𝑋 is the sum of independent Bernoulli(𝑝) variables and
each Bernoulli variable has variance 𝑝(1 − 𝑝) we have
𝑋 ∼ binomial(𝑛, 𝑝) ⇒ Var(𝑋) = 𝑛𝑝(1 − 𝑝).

Question 18

Bernoulli(p)

Distribution, range 𝑋, pmf 𝑝(𝑥), mean 𝐸[𝑋], variance Var(𝑋)

Answer 18

Question 19

Binomial(n, p)

Distribution, range 𝑋, pmf 𝑝(𝑥), mean 𝐸[𝑋], variance Var(𝑋)

Answer 19

Question 20

Uniform(n)

Distribution, range 𝑋, pmf 𝑝(𝑥), mean 𝐸[𝑋], variance Var(𝑋)

Answer 20

Question 21

Geometric(p)

Distribution, range 𝑋, pmf 𝑝(𝑥), mean 𝐸[𝑋], variance Var(𝑋)

Answer 21

Question 22

Expected Value
Let X be a discrete random variable with range

Answer 22

Question 23

Variance

Answer 23