Probability Distributions or Probability Histograms for Discrete Random Variables

For a complete description of discrete random variable X, there exists a probability distribution that simply tables the possible outcomes f(x) and their probabilities p(x):

Example: Tossing two fair coins where random variable X represents the number of heads observed. Imagine what can happen:

Toss 1:	Heads	Heads	Tails	Tails
Toss 2:	Heads	Tails	Heads	Tails

So random variable X can look like: 0 heads, 1 head, 1 head, 2 heads, these are the possible things that can happen. And so it looks like the outcome of getting 1 head happens twice as often as getting either two or no heads. You can table this discrete random variable in the following manner:

f(x)	0	1	2
p(X)	1/4	1/2	1/4

How did I figure out the probabilities? Well, the chance of getting a tail on the first toss is 1/2 and the chance of getting a tail on the second toss is also 1/2 so 1/2 * 1/2 = 1/4

The chance of getting a head on the first toss is 1/2 the chance of getting a tail on the second toss is 1/2 so (H,T) = 1/2* 1/2 = 1/4 but since I can also have the situation where (T,H) counts as one head, the chance there is 1/2*1/2 too or 1/4. We add those together (because you can either get HT or TH but not both simultaneously) and the total chance is 1/2. Finally, the chance of getting two heads in two tosses is: 1/2*1/2 = 1/4.

Note the probabilities sum to 1.0

How would three days on the stock exchange look where random variable X represents the number of "up days" observed where the probability of an up day is .60, the probability of a neutral day is .10 and the probability of a down day is .30:

Start by "listing the ways" let U=up and D=down, N=neutral then outcomes for 3 days might be:

UUU NNN DND

UUD NUN DDN

UDU NNU NND

DUU UNN NDN

UDD NUU DNN

DUD UNU UND

DDU UUN NUD

DDD NDD NDU

NDU DNU DUN

27 ways! (Don't worry, there is a smarter way to do this, remember, all we care about is the number of "up days" observed) If we're only interested in the "up days" there can only be 4 things that can happen in 3 days:

So combine the down and neutral (.1 + .3 = .4). Now it looks like a lopsided coin-toss, U=up and N=Not up:

UUU (.6 * .6 * .6) UNN (.6 * .4 * .4)

UUN (.6 * .6 * .4) NUN (.4 * .6* .4)

UNU (.6 * .4 * .6) NNU (.4 * .4 * .6)

NUU (.4 * .6 * .6) NNN (.4 * .4 * .4)

And then combine them to fill out this table:

f(x)	0	1	2	3
p(X)

Once you have the outcomes and probabilities, you can start answering questions about how the random variable behaves. Such as:

(a) for the coin problem, what's the chance of seeing at least one head? (1/2 + 1/4)

(b) for the coin problem, what's the chance of seeing less than 2 heads? (1/4 + 1/2)

(c) for the trading days problem, what's the chance of seeing at least 2 "up days" in a 3 day period?

(d) For the trading days problem, what's the chance of seeing at least 1 "down day" in a 3 days period?

For discrete random variables, sometimes you are simply given the probabilities and the outcomes. See problems 4.39 and 4.41 in your text. And sometimes you are not, see problem 4.42 and 4.43 in your text.