Finding the Mode of a Discrete Random Variable

The question asks for the mode of a discrete random variable X, which has a given probability mass function (PMF).

The probability mass function (PMF) is given by:

\( P(X=k) = (1 - p)^{k - 1} p \)

where \( k = 1, 2, 3, \dots \) and \( 0 < p \le 1 \).

The support of the random variable X is the set of positive integers \( \{1, 2, 3, \dots\} \).

Understanding the Mode

For a discrete random variable, the mode is the value (or values) in the support that has the highest probability. To find the mode, we need to find the value of \( k \) that maximizes \( P(X=k) \).

Analyzing the Probability Mass Function

Let's look at the probability \( P(X=k) \) for different values of \( k \):

• For \( k=1 \): \( P(X=1) = (1 - p)^{1 - 1} p = (1 - p)^0 p = 1 \cdot p = p \)
• For \( k=2 \): \( P(X=2) = (1 - p)^{2 - 1} p = (1 - p)^1 p = (1 - p)p \)
• For \( k=3 \): \( P(X=3) = (1 - p)^{3 - 1} p = (1 - p)^2 p \)
• For \( k=4 \): \( P(X=4) = (1 - p)^{4 - 1} p = (1 - p)^3 p \)
• And so on: \( P(X=k) = (1 - p)^{k - 1} p \)

We want to find the value of \( k \) (from \( \{1, 2, 3, \dots\} \)) that makes \( (1 - p)^{k - 1} p \) as large as possible.

Since \( p \) is a positive constant (\( 0 < p \le 1 \)), maximizing \( P(X=k) = (1 - p)^{k - 1} p \) is equivalent to maximizing the term \( (1 - p)^{k - 1} \).

Case 1: \( p = 1 \)

If \( p = 1 \), the PMF becomes:

\( P(X=k) = (1 - 1)^{k - 1} \cdot 1 = 0^{k - 1} \)

• For \( k=1 \): \( P(X=1) = 0^{1 - 1} = 0^0 \). By convention in probability (and often in mathematics), \( 0^0 \) is taken as 1 in this context. So, \( P(X=1) = 1 \).
• For \( k > 1 \): \( k-1 \ge 1 \), so \( P(X=k) = 0^{k - 1} = 0 \).

When \( p=1 \), we have \( P(X=1) = 1 \) and \( P(X=k) = 0 \) for all \( k > 1 \). The highest probability is 1, which occurs at \( k=1 \). So, the mode is 1.

Case 2: \( 0 < p < 1 \)

If \( 0 < p < 1 \), then \( 0 < 1 - p < 1 \). Let \( q = 1 - p \), so \( 0 < q < 1 \).

The term we want to maximize is \( (1 - p)^{k - 1} = q^{k - 1} \).

We are looking for the value of \( k \in \{1, 2, 3, \dots\} \) that maximizes \( q^{k-1} \).

Consider the values of \( q^{k-1} \) for increasing \( k \):

• For \( k=1 \): \( q^{1 - 1} = q^0 = 1 \)
• For \( k=2 \): \( q^{2 - 1} = q^1 = q \)
• For \( k=3 \): \( q^{3 - 1} = q^2 \)
• For \( k=4 \): \( q^{4 - 1} = q^3 \)

Since \( 0 < q < 1 \), raising \( q \) to a higher positive power results in a smaller number (e.g., \( q^2 = q \cdot q < q \) because \( q < 1 \)). Therefore, the sequence \( q^0, q^1, q^2, q^3, \dots \) is a strictly decreasing sequence: \( 1 > q > q^2 > q^3 > \dots \).

The maximum value of \( q^{k - 1} \) occurs when the exponent \( k - 1 \) is smallest. The smallest possible value for \( k \) is 1, which gives the smallest exponent \( k-1 = 1-1 = 0 \). So, \( q^{k-1} \) is maximum at \( k=1 \), where \( q^{k-1} = q^0 = 1 \).

Since \( (1 - p)^{k - 1} \) is maximum at \( k=1 \), the probability \( P(X=k) = (1 - p)^{k - 1} p \) is also maximum at \( k=1 \).

So, for \( 0 < p < 1 \), the mode is 1.

Conclusion on the Mode

In both cases (\( p=1 \) and \( 0 < p < 1 \)), the maximum probability occurs at \( k=1 \). Therefore, the mode of the discrete random variable X is 1.

This specific probability distribution is known as the Geometric distribution, defined as the number of Bernoulli trials required to get the first success, with support \( \{1, 2, 3, \dots\} \). For this form of the Geometric distribution, the mode is always 1, regardless of the value of \( p \) (as long as \( 0 < p \le 1 \)).

Checking the Options

Let's compare our result with the given options:

1. 1: This matches our calculated mode.
2. 0: This value is not in the support of X (\(k\) must be 1, 2, 3, ...).
3. p: The mode must be a possible value of X (a value of \(k\)), not the parameter \(p\).
4. 1 - p: The mode must be a possible value of X (a value of \(k\)), not the parameter \(1-p\).

Based on our analysis, the correct mode is 1.

Revision Table: Key Concepts

Additional Information: Geometric Distribution

The probability mass function \( P(X=k) = (1 - p)^{k - 1} p \) for \( k = 1, 2, 3, \dots \) and \( 0 < p \le 1 \) corresponds to a Geometric distribution. Specifically, this form models the number of Bernoulli trials needed to get the first success, where \( p \) is the probability of success on a single trial.

• Expected Value: \( E[X] = \frac{1}{p} \)
• Variance: \( Var(X) = \frac{1-p}{p^2} \)
• Memoryless Property: \( P(X > m + n | X > m) = P(X > n) \) for non-negative integers \( m \) and \( n \).

Another common definition of the Geometric distribution has the support \( \{0, 1, 2, \dots\} \) and PMF \( P(Y=k) = (1-p)^k p \). This models the number of failures before the first success. For this alternative form, the mode is 0 if \( 0 < p < 1 \), and 0 (or any non-negative integer) if \( p=1 \). However, the question specifically provides the PMF and support \( k = 1, 2, 3, \dots \), which matches the first-success definition, where the mode is consistently 1.