Question

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

Answer 1

PMF for discrete random variable `X:" "` `p_X(x)" "` or `" "p(x)`.
Mean: `" "mu=E[X]=sum_x x*p(x)`.
Variance: `" "sigma^2 = "Var"[X]=sum_x [x^2*p(x)] - [sum_x x*p(x)]^2`.

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 `X` is the result of a single dice roll, then `X` could take on the values `{1,2,3,4,5,6},` each with equal probability `1/6`. The pmf for `X` would be:

`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: `p(x)` is equal to `P(X=x)`.

The mean `mu` (or expected value `E[X]`) of a random variable `X` is the sum of the weighted possible values for `X`; weighted, that is, by their respective probabilities. If `S` is the set of all possible values for `X`, then the formula for the mean is:

`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` (or `"Var"[X]`) of a random variable `X` is a measure of the spread of the possible values. By definition, it is the expected value of the squared distance between `X` and `mu`:

`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 `mu" "(E[X]),` so now we just need a formula for `E[X^2].` This is the expected value of the squared random variable, so our formula for this is the sum of the squared possible values for `X`, again, weighted by the probabilities of the `x`-values:

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

Using this, our formula for the variance of `X` becomes

`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, `mu` was calculated to be `3.5,` so we use that for our last term to get

`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`