Answered

Which of the following represents a valid probability distribution?



\multicolumn{2}{|r|}{Probability Distribution A}

$X$		$P(x)$

0		0

1		0

2		0

3		1

Verified Answer :

  • A probability distribution must satisfy \(0 \le P(X) \le 1\) for all \(X\) and \(\sum P(X) = 1\).
  • The provided table gives \(P(X=0)=0\), \(P(X=1)=0\), \(P(X=2)=0\), and \(P(X=3)=1\).
  • Checking the conditions:
  • All \(P(X) \ge 0\) and \(P(X) \le 1\).
  • The sum \(\sum P(X) = 0+0+0+1 = 1\).
  • Since both conditions are satisfied, the table \(\boxed{\text{is a valid probability distribution}}\).

Explanation

  1. Understand the requirements for a valid probability distribution

To determine if a given table represents a valid probability distribution, we must evaluate it against two fundamental criteria that define such distributions for a discrete random variable \(X\):

  1. Each individual probability \(P(X=x)\) must be between 0 and 1 inclusive, i.e., \(0 \le P(X=x) \le 1\).
  2. The sum of all probabilities over the entire sample space must equal exactly 1, i.e., \(\sum P(X=x) = 1\).
  1. Evaluate the provided distribution table

Now, let's examine "Probability Distribution A" provided in the table:

  • For \(X=0\), \(P(X=0) = 0\).
  • For \(X=1\), \(P(X=1) = 0\).
  • For \(X=2\), \(P(X=2) = 0\).
  • For \(X=3\), \(P(X=3) = 1\).

First, we check condition 1: All values (\(0, 0, 0, 1\)) are within the range \([0, 1]\). This condition is satisfied.
Next, we check condition 2 by calculating the sum: \(S = 0 + 0 + 0 + 1 = 1\). Since the sum is exactly \(1\), this condition is also satisfied.

  1. Conclusion about the distribution

Since both necessary conditions are met—each probability is between 0 and 1, and the total sum of the probabilities is equal to 1—we conclude that the table represents a valid probability distribution. This is a degenerate distribution where the variable \(X\) is certain to take the value 3.

Examples

In statistical modeling, a degenerate distribution is a useful concept to represent a constant value as a random variable. For example, if a machine is guaranteed to produce an item with a weight of exactly 3 kg, we can model the weight \(W\) as a random variable where \(P(W=3) = 1\) and \(P(W eq 3) = 0\), which is a valid probability distribution that reflects perfect consistency in production.