Question

If the probability function of a random variable \(X\) is \[ P(X=x)=ak^x,\qquad x=0,1,2,\ldots, \] then the value of \(k\) is

• A. \(1-a,\;0<a<1\)
• B. \(1-a\) for all positive \(a\)
• C. \(\dfrac{1}{1-a}\)
• D. \(a-1\)

Hint:

Whenever a probability mass function contains terms like \(a,ak,ak^2,\ldots\), immediately think of the infinite geometric series formula \[ 1+r+r^2+\cdots=\frac1{1-r}. \]

Answer 1

The Correct Option is A

Concept: For a probability mass function, \[ \sum P(X=x)=1. \] Since \(X\) takes values \[ 0,1,2,\ldots, \] the given distribution forms an infinite geometric series.

Step 1: Apply the total probability condition \[ \sum_{x=0}^{\infty}P(X=x)=1. \] Substituting, \[ \sum_{x=0}^{\infty}ak^x=1. \] Taking \(a\) common, \[ a\sum_{x=0}^{\infty}k^x=1. \]

Step 2: Use the geometric series formula For \[ |k|<1, \] \[ \sum_{x=0}^{\infty}k^x = \frac{1}{1-k}. \] Therefore, \[ a\left(\frac1{1-k}\right)=1. \] \[ a=1-k. \] Hence, \[ k=1-a. \]

Step 3: Determine the restriction Since the infinite geometric series must converge, \[ |k|<1. \] Also probabilities must be positive, therefore \[ 0<a<1. \] Thus, \[ \boxed{k=1-a,\quad 0<a<1}. \] Hence the correct answer is \[ \boxed{(A)}. \]