How do you use a probability mass function to calculate the mean and variance of a discrete distribution?
1 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
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
#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
#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#