Validate, complete and normalise a discrete probability distribution

Which two checks come first when deciding whether a table is a discrete probability distribution?

Vary q. The total stays at 1; inspect each entry as well. Decimal table entries represent the exact fractions in this example

Which settings give a valid distribution?

A random variable and one of its values

A random variable assigns a number to each outcome of an experiment. X names the random variable; x names one numerical value. Different outcomes may give the same value. A negative value of X is allowed: it is the probability, not the value, that must lie between 0 and 1.

Properties of a discrete probability distribution

A discrete probability distribution is valid only when every probability is between 0 and 1 inclusive and all probabilities add to 1. A mass-function rule is used only on its stated discrete value set

1. 1Use the stated value set Evaluate a mass function only at the listed discrete values \(\displaystyle x\in\{-1,0,2\}\)
2. 2Check every probability Each probability must be at least 0 and at most 1. \(\displaystyle 0\leq\frac{3}{8}\leq1\)
3. 3Check the total Add all probabilities for the complete value set \(\displaystyle \frac{1}{8}+\frac{1}{4}+\frac{5}{8}=1\)
4. 4Complete a missing entry Subtract the known total from 1 to find one missing probability \(\displaystyle p_3=1-\left(\frac{1}{4}+\frac{3}{8}\right)=\frac{3}{8}\)

Listed probabilities

\(\displaystyle \frac14,\quad\frac18,\quad\frac58\)

For probabilities 1/4, 1/8 and 5/8, add the first two entries

Listed probabilities

\(\displaystyle \frac14,\quad\frac18,\quad\frac58\)

Add the three listed probabilities

Accepting a total without checking bounds. One of these lines is right. Which one?

Distribution data

\(\displaystyle q,\quad\frac14,\quad\frac34-q\)

Find the largest q that keeps every listed probability valid

Is a zero entry allowed in a displayed probability table?

Normalise over the stated discrete set

For this proposed rule, evaluate x+3 at −2, 0 and 3, then add those three weights. Do not integrate over the interval or include unlisted intermediate inputs

Distribution data

\(\displaystyle p(x)=k(x+3),\quad x\in\{-2,0,3\}\)

Find k so that the three probabilities sum to 1.

Distribution data

\(\displaystyle p(x)=\frac{x+3}{10},\quad x\in\{-2,0,3\}\)

Evaluate the normalised mass function at x=3.

Check all entries, not just the total equation

The probabilities are nonnegative, at most one and sum to one. On a stick chart they are heights at the three listed values, not a continuous curve with an area interpretation

The complete value set contains minus two, zero and three. What is the probability that X equals one for this valid distribution?

Distribution data

\(\displaystyle \Pr(X=-1)=\frac16,\quad\Pr(X=2)=\frac5{12},\quad\Pr(X=5)=a\)

Complete the third probability

Use one total equation for the parameter

These three expressions are proposed probabilities for the complete value set. Add all three, set the total equal to 1, solve, and substitute into each original expression

Distribution data

\(\displaystyle 2k+(k+\frac18)+(\frac38-k)=1\)

Find the normalisation constant k.

Distribution data

\(\displaystyle \Pr(X=1)=k+\frac18,\quad k=\frac14\)

Use k=1/4 to find Pr(X=1)

Return to every original probability

The total equation is necessary, but each resulting entry must also be admissible. Here the three substituted values pass both checks

Keep both algebraic roots before testing them

The probability expressions may produce a quadratic total equation. Find both roots first, then use the probability bounds to decide which root is admissible. Enter root lists in increasing order when that order is requested

Form and factor the total equation

Adding the original three expressions gives the total equation shown. Clearing the denominator and factoring gives two algebraic candidates; neither is accepted solely because it solves the total equation

Distribution data

\(\displaystyle (2k-1)(k+1)=0\)

Find the smaller algebraic root

Distribution data

\(\displaystyle (2k-1)(k+1)=0\)

Find the larger algebraic root

Distribution data

\(\displaystyle \Pr(X=-1)=k^2+\frac18,\quad k=-1\)

At k=−1, calculate Pr(X=−1)

Reject the probability values, not merely a sign

For the first root, the displayed entries sum to one but violate the probability bounds. For the second root, every entry is valid. Thus only the second root is admissible

Why should a negative parameter not be rejected automatically?

Construct a distribution and justify it, ungraded

On paper, find c, construct the complete probability table and verify both defining properties. Draw a stick chart at the listed values without joining them into a continuous curve. State the probability at an unlisted value, explaining the restriction on the formula

Distribution data

\(\displaystyle p(x)=c(2x+3),\quad x\in\{-1,2,4\}\)

After constructing the full table, enter c.

Distribution data

\(\displaystyle p(x)=\frac{2x+3}{19},\quad x\in\{-1,2,4\}\)

Enter the probability at x=2.

Check the whole distribution

The three exact probabilities are nonnegative and sum to 1. Draw sticks at x=−1, 2 and 4 with these heights. There is no mass at x=0, so its probability is zero; do not substitute 0 into the branch formula

Use the value set and both probability conditions

A distribution assigns mass to discrete values. Complete a missing entry from the total, evaluate a rule only on its stated set, and test every normalisation root in the original probabilities. A total of one alone does not establish validity