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

Step 1: Use the total probability rule.
For any probability distribution, all the probabilities must add up to $1$. So $\sum_{x=0}^{\infty}P(X=x)=1$.

Step 2: Write the sum.
Here $P(X=x)=ak^x$ for $x=0,1,2,\ldots$. So \[ \sum_{x=0}^{\infty}ak^x=1. \]

Step 3: Take out the constant.
Pull $a$ outside the sum: \[ a\sum_{x=0}^{\infty}k^x=1. \]

Step 4: Sum the geometric series.
When $|k|<1$, the geometric series gives $\sum_{x=0}^{\infty}k^x=\dfrac{1}{1-k}$.

Step 5: Solve the equation.
So $a\cdot\dfrac{1}{1-k}=1$, which means $a=1-k$, and therefore $k=1-a$.

Step 6: Note the restriction.
For the series to converge we need $|k|<1$, which forces $0<a<1$ when probabilities are positive.

Step 7: State the answer.
So $k=1-a$, with $0<a<1$. \[ \boxed{k=1-a,\ 0<a<1} \]