jhenezhubby01ff

2022-10-09

Suppose U and V are independent and follow the geometric distribution.

$p(k)=p(1-p{)}^{k}\text{}$ for $\text{}k=0,1,...$

Define the random variable $Z=U+V$

(a) Determine the joint probability mass function ${P}_{\text{}U,Z}(u,z)=Pr(U=u,\text{}Z=z)$

(b) Determine the conditional probability mass function for U given that $Z=n$

$p(k)=p(1-p{)}^{k}\text{}$ for $\text{}k=0,1,...$

Define the random variable $Z=U+V$

(a) Determine the joint probability mass function ${P}_{\text{}U,Z}(u,z)=Pr(U=u,\text{}Z=z)$

(b) Determine the conditional probability mass function for U given that $Z=n$

Nadia Berry

Beginner2022-10-10Added 7 answers

Step 1

$\begin{array}{rl}Pr(U=u)=& Pr(\text{number of failures before the first success}=u)\\ =& Pr(\text{failure on the first}u\text{trials and success on trial}\mathrm{\#}(u+1))\\ =& p(1-p{)}^{u}.\\ Pr(Z=n)=& Pr{\textstyle (}(U=0\text{}\mathrm{}\text{}V=n)\text{or}(U=1\text{}\mathrm{}\text{}V=n-1)\\ & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\text{or}(U=2\text{}\mathrm{}\text{}V=n-2)\text{or}(U=3\text{}\mathrm{}\text{}V=n-3)\\ & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\text{or}(U=4\text{}\mathrm{}V=n-4)\text{or}\cdots \cdots \cdots {\textstyle )}\\ =& p\cdot {\textstyle (}p(1-p{)}^{n}{\textstyle )}+{\textstyle (}p(1-p){\textstyle )}\cdot {\textstyle (}p(1-p{)}^{n-1}{\textstyle )}+{\textstyle (}p(1-p{)}^{2}{\textstyle )}\cdot {\textstyle (}p(1-p{)}^{n-2}{\textstyle )}\\ & +{\textstyle (}p(1-p{)}^{3}{\textstyle )}\cdot {\textstyle (}p(1-p{)}^{n-3}{\textstyle )}+\cdots \cdots \\ =& (n+1){p}^{2}(1-p{)}^{n}.\\ \end{array}$

Step 2

$\begin{array}{rl}Pr(U=u\mid Z=n)& =\frac{Pr(U=u\text{}\mathrm{}\text{}Z=n)}{Pr(Z=n)}=\frac{Pr(U=u\text{}\mathrm{}\text{}V=n-u)}{Pr(Z=n)}\\ & =\frac{p(1-p{)}^{u}\cdot p(1-p{)}^{n-u}}{(n+1){p}^{2}(1-p{)}^{n}}=\frac{1}{n+1}.\end{array}$

Thus all $n+1$ possible values of U are equally probable.

$\begin{array}{rl}Pr(U=u)=& Pr(\text{number of failures before the first success}=u)\\ =& Pr(\text{failure on the first}u\text{trials and success on trial}\mathrm{\#}(u+1))\\ =& p(1-p{)}^{u}.\\ Pr(Z=n)=& Pr{\textstyle (}(U=0\text{}\mathrm{}\text{}V=n)\text{or}(U=1\text{}\mathrm{}\text{}V=n-1)\\ & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\text{or}(U=2\text{}\mathrm{}\text{}V=n-2)\text{or}(U=3\text{}\mathrm{}\text{}V=n-3)\\ & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\text{or}(U=4\text{}\mathrm{}V=n-4)\text{or}\cdots \cdots \cdots {\textstyle )}\\ =& p\cdot {\textstyle (}p(1-p{)}^{n}{\textstyle )}+{\textstyle (}p(1-p){\textstyle )}\cdot {\textstyle (}p(1-p{)}^{n-1}{\textstyle )}+{\textstyle (}p(1-p{)}^{2}{\textstyle )}\cdot {\textstyle (}p(1-p{)}^{n-2}{\textstyle )}\\ & +{\textstyle (}p(1-p{)}^{3}{\textstyle )}\cdot {\textstyle (}p(1-p{)}^{n-3}{\textstyle )}+\cdots \cdots \\ =& (n+1){p}^{2}(1-p{)}^{n}.\\ \end{array}$

Step 2

$\begin{array}{rl}Pr(U=u\mid Z=n)& =\frac{Pr(U=u\text{}\mathrm{}\text{}Z=n)}{Pr(Z=n)}=\frac{Pr(U=u\text{}\mathrm{}\text{}V=n-u)}{Pr(Z=n)}\\ & =\frac{p(1-p{)}^{u}\cdot p(1-p{)}^{n-u}}{(n+1){p}^{2}(1-p{)}^{n}}=\frac{1}{n+1}.\end{array}$

Thus all $n+1$ possible values of U are equally probable.

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