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

Short 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-by-step solution

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$

: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$