Question

For each of the following, find the constant \(c\) so that \(p(x)\) satisfies the condition of being a pmf of one random variable \(X\). (a) \(p(x)=c\left(\frac{2}{3}\right)^{x}, x=1,2,3, \ldots\), zero elsewhere. (b) \(p(x)=c x, x=1,2,3,4,5,6\), zero elsewhere.

Short answer

For problem (a), the constant \(c=\frac{3}{2}\) and for problem (b), the constant \(c=\frac{2}{21}\).

Step-by-step solution

Solve problem (a)

First, consider \(p(x)=c\left(\frac{2}{3}\right)^{x}, x=1,2,3, \ldots\). Since this is a probability mass function (pmf), the total probability should sum up to 1. So, find the summation of probabilities from \(x=1\) to infinity and equate it to 1: \(\sum_{{x=1}}^{\infty}c\left(\frac{2}{3}\right)^{x}=1\). This is a geometric series where \(r=\frac{2}{3}\), so its sum is given by \(\frac{a}{1-r}=1\) where \(a=c\frac{2}{3}\). After some calculation, we find \(c=\frac{3}{2}\).

Solve problem (b)

Next, consider \(p(x)=c x, x=1,2,3,4,5,6\). Again, since this is a pmf, the total probability should sum up to 1: \(\sum_{{x=1}}^{6}c x=1\). This forms an arithmetic series. So, we can calculate its sum as \(\frac{n}{2}(a+l) = 1\), where \(n=6\), \(a=1\), and \(l=6\). After some calculation, we find \(c=\frac{2}{21}\).