Question

Let \(X\) be a random variable such that the probability function of a distribution is given by \(P(X=0)=\frac12,\ P(X=j)=\frac{1}{3^j}\ (j=1,2,3,\dots,\infty)\). Then the mean of the distribution and \(P(X\) is positive and even\()\) respectively are :

• A. \(\frac38\) and \(\frac18\)
• B. \(\frac34\) and \(\frac18\)
• C. \(\frac34\) and \(\frac19\)
• D. \(\frac34\) and \(\frac1{16}\)

Hint:

Mean \(=\sum j/3^j\) (arithmetico-geometric series); even \(j\) give a geometric series with ratio \(1/9\).

Answer 1

The Correct Option is B

Check: \(\frac12+\sum_{j\ge1}\frac1{3^j}=\frac12+\frac{1/3}{1-1/3}=1\).
Mean \(S=\sum_{j\ge1}\frac{j}{3^j}\). Then \(S-\frac S3=\sum_{j\ge1}\frac1{3^j}=\frac12\Rightarrow\frac23S=\frac12\Rightarrow S=\frac34\).
\(P(X\text{ positive even})=\frac1{3^2}+\frac1{3^4}+\dots=\frac{1/9}{1-1/9}=\frac18\).
Answer: (2) \(\frac34\) and \(\frac18\)