Q.4.4 - A First Course in Probability

Question

The random variable X is said to have the Yule-Simons distribution if 

$P\{X=n\}=\frac{4}{n(n+1)(n+2)},n\ge 1$

(a) Show that the preceding is actually a probability mass function. That is, show that $\sum _{n=1}^{\infty }P\{X=n\}=1$

(b) Show that E[X] = 2. 

(c) Show that E[X2] = q 

Step-by-Step Solution

Answer

In the given information the answer of part(a) is$\sum _{n=1}^{+\infty }P(X=n)=1$

part(b) is $E\left(X\right)=2$

part (c) is $E\left(X^2\right)=+\infty$

Step 1:Given Information (Part-a)

Given that $P(X=n)=\frac{4}{n(n+1)(n+2)},n\ge 1$.

Step 2:Calculation (Part-a)

$\sum _{n=1}^{+\infty }P(X=n)$

                $=\sum _{n=1}^{+\infty }\frac{4}{n(n+1)(n+2)}$

                 $=\sum _{n=1}^{+\infty }\left(\frac{4}{n(n+1)}-\frac{4}{n(n+2)}\right)$

                  $=\sum _{n=1}^{+\infty }\left(\frac{4}{n}-\frac{4}{n+1}-\frac{4}{n(n+2)}\right)$

                 $=\sum _{n=1}^{+\infty }\left(\frac{4}{n}-\frac{4}{n+1}-\frac{2}{n}+\frac{2}{n+2}\right)$

                   $=2+1+2\sum _{n=3}^{+\infty }\frac{1}{n}-2-4\sum _{n=3}^{+\infty }\frac{1}{n}+2\sum _{n=3}^{+\infty }\frac{1}{n}$

                   $=1$

                  

Step 3:Final Answer ( Part-a)

The answer is $\sum _{n=1}^{+\infty }P\left(X=n\right)=$$1$

Step4 :Given Information (Part-b)

Given that $P(X=n)=\frac{4}{n(n+1)(n+2)},n\ge 1$

The expected value is the sum of each possibility n, with its possibility .

Step 5:Calculation(Part-b)

$E\left(X\right)=\sum _{i=1}^{+\infty }nP(x=n)$

         $=\sum _{n=1}^{+\infty }\frac{4n}{n(n+1)(n+2)}$

          $=\sum _{n=1}^{+\infty }\frac{4}{(n+1)(n+2)}$

           $=4\sum _{n=2}^{+\infty }\frac{1}{n}-4\sum _{n=3}^{+\infty }\frac{1}{n}$

             = $4\times \frac{1}{2}$

              $=2$

Step 6:Final Answer (Part-b)

The answer is $E\left(X\right)=2$

Step 7:Given Information (Part-c)

Given that $P(X=n)=\frac{4}{n(n+1)(n+2)},n\ge 1$

Step 8:Calculation (Part-c)

$E\left(X^2\right)=\sum _{i=1}^{+\infty }{n}^{2}P(x=n)$

          $=\sum _{n=1}^{+\infty }\frac{4n}{n(n+1)(n+2)}$

           $=\sum _{n=1}^{+\infty }\frac{4n}{(n+1)(n+2)}$

            $=2+4\sum _{n=3}\frac{1}{n}$

              $=2+4(+\infty )$

             =  $+\infty$

Step 9:Final Answer (Part-c)

The  answer is$E\left(X^2\right)=$$+\infty$