Distribution of a finite random variable

When several ordered outcomes give the same value of X, what should happen first?

Vary the common success probability p for two independent trials. Compare the three count probabilities; having three listed count values does not make their probabilities equal

At p=1/2, why is a count of one more likely than a count of zero?

A finite random-variable distribution

A finite random variable assigns a numerical value to each elementary outcome. Group all outcomes with the same value and add their exact probabilities to form the distribution. The binary numerical illustration here uses two independent fair trials, so its four ordered outcomes are equally likely

1. 1Define X by a rule Apply the stated rule to each elementary outcome \(\displaystyle X(\text{success},\text{failure})=1\)
2. 2Keep ordered outcomes Outcomes in different stage orders remain different elementary outcomes \(\displaystyle (A,B)\ne(B,A)\)
3. 3Group equal values Add the masses of every outcome that produces the same value \(\displaystyle \Pr(X=1)=\Pr(S,F)+\Pr(F,S)=\frac{2}{4}\)
4. 4Check the total mass The probabilities of all possible X values must add to one \(\displaystyle \sum_x\Pr(X=x)=\frac{1}{4}+\frac{1}{2}+\frac{1}{4}=1\)

Two independent fair binary trials

\(\displaystyle X=\text{number of successes},\quad\omega=(\text{success},\text{success})\)

For X equal to the number of successes, calculate X for the ordered outcome (success, success)

Two independent fair binary trials

\(\displaystyle S=\{(S,S),(S,F),(F,S),(F,F)\},\quad\Pr(\omega)=\frac14,\quad X=\text{number of successes}\)

Use the four equally likely ordered outcomes to find Pr(X = 2)

For two independent fair binary trials, let X be the number of successes. Each ordered outcome has probability one quarter. Which line is right?

Keep zero-mass entries distinct from possible values

At p=0, only count zero can occur; at p=1, only count two can occur. The explorer still lists the three candidate counts for comparison. The effective support is the subset with positive probability

Construct a distribution from stated joint masses

Stage one records 1 or 3 and stage two records 0 or 2. The four ordered pairs have the probabilities shown, and X is their code sum. Independence is not given: use these joint masses directly

Experiment and rule

\(\displaystyle X(a,b)=a+b\)

Find the value of X for the ordered outcome (3,0)

X is the sum of the two codes in an ordered pair. Which outcomes must be grouped together when finding Pr(X=3)?

Experiment and rule

\(\displaystyle S=\{(1,0),(1,2),(3,0),(3,2)\},\quad\Pr(1,0)=\frac1{12},\quad X(a,b)=a+b\)

Find Pr(X=1)

Experiment and rule

\(\displaystyle S=\{(1,0),(1,2),(3,0),(3,2)\},\quad X(a,b)=a+b,\quad\Pr(1,2)=\frac14,\quad\Pr(3,0)=\frac13\)

Find Pr(X=3)

Experiment and rule

\(\displaystyle S=\{(1,0),(1,2),(3,0),(3,2)\},\quad\Pr(3,2)=\frac13,\quad X(a,b)=a+b\)

Find Pr(X=5)

Check the complete grouped distribution

The masses total one. Grouping reduces four ordered outcomes to three values, but it does not make those values equiprobable

Experiment and rule

\(\displaystyle X\in\{1,3,5\},\quad\Pr(X=3)=\frac7{12},\quad\Pr(X=5)=\frac13\)

Find Pr(X≥3) from the grouped distribution

Enumerate two independent binary trials

Each trial succeeds with probability 1/4 and fails with probability 3/4. Independence allows the two stage probabilities to be multiplied for each ordered pair. Count successes after forming the four pairs; no general binomial formula is needed

Experiment and rule

\(\displaystyle \Pr(F)=\frac34,\quad\text{trials independent}\)

Find the probability of zero successes

Experiment and rule

\(\displaystyle \Pr(SF)=\frac3{16},\quad\Pr(FS)=\frac3{16}\)

Find the probability of exactly one success

Experiment and rule

\(\displaystyle \Pr(S)=\frac14,\quad\text{trials independent}\)

Find the probability of two successes

Check enumeration before using a count distribution

The displayed probabilities for counts zero, one and two total one. The middle value combines two elementary outcomes. A count ratio would be invalid because these elementary masses are unequal

Equal marginal probabilities do not establish independence

In a separate experiment, the joint probabilities are shown in the table. They total one. This table, rather than an independence assumption, determines the distribution

Experiment and rule

\(\displaystyle \Pr(SS)=\frac25,\quad\Pr(SF)=\frac1{10}\)

Find the probability that the first stage succeeds in this separate experiment

Experiment and rule

\(\displaystyle \Pr(SF)=\frac1{10},\quad\Pr(FS)=\frac1{10}\)

Find the probability of exactly one success in this separate experiment

Check independence against the actual joint table

The probability of SS is 2/5, whereas the product of the two marginal success probabilities is 1/4. These trials are not independent. Their success-count distribution is 2/5,1/5,2/5.

Construct two complete distributions, ungraded

The ordered pairs (0,0),(0,2),(2,0),(2,2) have masses 1/10,3/10,1/5,2/5. Write every pair and both mapped values. Form the full distributions for Y, the maximum, and Z, the minimum, and check each total. The component entries do not grade the complete mapping tables or explanation

Experiment and rule

\(\displaystyle S=\{(0,0),(0,2),(2,0),(2,2)\},\quad Y(a,b)=\max(a,b),\quad\Pr(0,2)=\frac3{10},\quad\Pr(2,0)=\frac15,\quad\Pr(2,2)=\frac25\)

After completing the full mapping, enter Pr(Y=2)

Experiment and rule

\(\displaystyle S=\{(0,0),(0,2),(2,0),(2,2)\},\quad\Pr(2,2)=\frac25,\quad Z(a,b)=\min(a,b)\)

Enter Pr(Z=2) for the same experiment

Compare the complete mappings and totals

The maximum maps the pairs to 0,2,2,2; the minimum maps them to 0,0,0,2. Both use the same joint masses but group them differently. Written mapping and distribution tables remain self-reviewed

Separate the outcome, its value and its probability

Enumerate the complete ordered sample space, apply the given rule, and add all masses in each value group. Use stated joint probabilities directly unless independence is supplied. Check non-negative masses and total one, then add the required values for an event