Random Variables and Distributions

Example 1: We roll a fair die, X is the number shown on the die
Example 2: We roll two fair dice, X is the sum of the dice
Example 3: We roll two fair dice, X is the absolute difference of the dice
Example 4: We roll two fair dice, X is the number of "sixes"
Example 5: We randomly choose 10 employees of WRInc, X is the number of employees in the sample with a job level of 1

Discrete Random Variables

Example 1 above: X takes one of the values {1,2,3,4,5,6}, so if we had 1000 observations drawn from this rv it would be only these numbers, repeated many times.
Example 6: We repeatedly roll a fair die. X is the number of rolls needed until the first "six", then X takes one of the values {1,2,3, ...}
Here theoretically X can take any (integer) value, but in practise we will only ever see a few of them (1-10 or so), and these would repeat many times.

The probability mass function (pmf) of a discrete random variable X is a list of the values that X takes and their respective probabilities.

Example 1 above:

x	1	2	3	4	5	6
P(x)	1/6	1/6	1/6	1/6	1/6	1/6

Example 4 above:

x	0	1	2
P(x)	25/36	10/36	1/36

Notice: the probabilities always sum up to 1!

Example We randomly select 2 employees of WR Inc. Let the random variable X be the nuber of females selected. Find the pmf of X.

What are the possible values of X? 0, 1 and 2 (none, exactly one and two females). What are the probabilities of these outcomes?

P(X=0) = P(both people selected are male) = P(first is male and second is male) = P(first is male)P(second is male|first is male) = 321/527·320/526 = 0.3706
P(X=2) = P(both people selected are female) = P(first is female and second is female) = P(first is female)P(second is female|first is female) = 206/527·205/526 = 0.1523

So we get the following table:

x	0	1	2
P(x)	0.3706		0.1523

but the probabilities have to sum up to 1, so the one missing is
P(X=1)=1-0.3706-0.1523=0.4771 and we have

x	0	1	2
P(x)	0.3706	0.4771	0.1523

Example 6 above: here we have infinitely many outcomes, so we need to write the pmf a little different, namely using an equation instead of a table. First let A_i="no six on i^th roll. Then we find

P(X=1) = P(six on first roll) = P(A^c₁) = 1/6

P(X=2) = P(no six on first roll, six on second roll) =
P(A₁A^c₂) =
P(A₁)P(A^c₂) (by independence)
= 5/6·1/6

P(X=3) = P(A₁A₂A^c₃) =
P(A₁)P(A₂)P(A^c₃) (by independence)
= 5/6·5/6·1/6

It's easy to guess the general case:

P(X=k) = (5/6)^k-1×1/6, k=1,2,..
So for example
P(first six on the 5^th roll) = P(X=5) = (5/6)^5-1×1/6 = 0.0804

Continuous Random Variables