Question

Which of the following could be a valid PMF of a random variable X ? Select one:

a. P(X=k)=(2ᵏ⁻¹)/(2ⁿ⁻¹)​∋k∈{1,2,…,n}
b. P(X=k)=1/k∋k∈{2,3,4,…}
c. P(X=k)=1/k(k+1)​∋k∈{1,2,…,n}
d. P(X=k)=1/2ᵏ​∋k∈{1,2,…,n}

Answer 1

The valid PMF of a random variable X among the given options is c. P(X=k)=1/k(k+1) for k∈{1,2,…,n}, as it satisfies both conditions required for a PMF.

Step-by-step explanation:

Valid Probability Mass Function (PMF)

To identify a valid PMF for a random variable X, two conditions must be satisfied:

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Let's evaluate the options given:

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Therefore, the valid PMF of a random variable X is option c.

Answer 2

Option c, P(X=k) = 1/k(k+1), is the potentially valid PMF for a random variable X, as it represents a telescoping series that approaches a sum of 1.

Step-by-step explanation:

To determine which option could be a valid probability mass function (PMF), we need to check two conditions for PMFs: first, the probability P(X=k) must be non-negative for each k in the domain, and second, the sum of all probabilities over the domain must equal 1. Now let's analyze each option provided.

• Option a: P(X=k) = (2k-1)/(2n-1) involves an exponential numerator which may grow faster than the denominator, thus not guaranteeing a sum of 1.
• Option b: P(X=k) = 1/k is an infinite series known to diverge (the harmonic series), so it does not sum up to 1.
• Option c: P(X=k) = 1/k(k+1) for k=1,2,...,n is a telescoping series where each term is the difference of the fractions 1/k and 1/(k+1), and so the sum from k=1 to n equals 1-1/(n+1), which approaches 1 as n goes to infinity. Therefore, for a finite n, the sum is less than 1 but can be made equal to 1 by adjusting the last term, making option c a valid PMF.
• Option d: P(X=k) = 1/2k is a convergent geometric series but does not necessarily sum to 1 for any finite n.

Hence, the potentially valid PMF of the random variable X among the given options is option c.