# How do you use a probability mass function to calculate the mean and variance of a discrete distribution?

##### 1 Answer

#### Answer:

PMF for discrete random variable

Mean:

Variance:

#### Explanation:

The probability mass function (or pmf, for short) is a mapping, that takes all the possible discrete values a random variable could take on, and maps them to their probabilities. Quick example: if

#p_X(x)={(1/6",", x in {1,2,3,4,5,6}),(0",","otherwise"):}#

*If we're only working with one random variable, the subscript #X# is often left out, so we write the pmf as #p(x)#.*

In short:

The **mean**

#mu =sum_(x in S) x*p(x)# .

In our example from above, this works out to be

#mu = sum_(x=1)^6 x*p(x)#

#color(white)mu = 1(1/6)+2(1/6)+3(1/6)+...+6(1/6)#

#color(white)mu = 1/6(1+2+3+4+5+6)#

#color(white)mu = 1/6(21)#

#color(white)mu = 3.5#

The **variance** *spread* of the possible values. By definition, it is the expected value of the squared distance between

#sigma^2 = E[(X-mu)^2]#

With some simple algebra and probability theory, this becomes

#sigma^2 = E[X^2] - mu^2#

We already have a formula for *squared* random variable, so our formula for this is the sum of the *squared* possible values for

#E[X^2]=sum_(x in S) x^2*p(x)#

Using this, our formula for the variance of

#sigma^2 =sum_(x in S) [x^2*p(x)] - mu^2#

#color(white)(sigma^2) =sum_(x in S) [x^2*p(x)] - [sum_(x in S) x*p(x)]^2#

For our example,

#sigma^2 =sum_(x=1)^6 [x^2*p(x)] - mu^2#

#color(white)(sigma^2) =[1^2(1/6)+2^2(1/6)+...+6^2(1/6)] - (3.5)^2#

#color(white)(sigma^2) =1/6(1+4+9+16+25+36)" "-" "(3.5)^2#

#color(white)(sigma^2) =1/6(91)" "-" "12.25#

#color(white)(sigma^2) ~~ 15.167-12.25#

#color(white)(sigma^2) = 2.917#