Jaxon Hamilton

2022-07-18

X and Y are geometric RV's with parameter p.

A) $P\{X+Y=n\}(n=1,2,...)$?

A) $P\{X+Y=n\}(n=1,2,...)$?

Cael Cox

Beginner2022-07-19Added 11 answers

Step 1

I am assuming X and Y are independent. Without this assumption, it is not possible to answer your question without further information.

Step 2

So, you have to be a little more clever: Suppose we let $Z=X+Y$. We wish to find $Pr[Z=n]$. Note that we can write $Pr[Z=n]=\sum _{y=0}^{n}Pr[(X=n-y)\cap (Y=y)]=\sum _{y=0}^{n}Pr[X=n-y]Pr[Y=y].$

Now you know what each of these probabilities is, so write them out, and calculate the sum.

I am assuming X and Y are independent. Without this assumption, it is not possible to answer your question without further information.

Step 2

So, you have to be a little more clever: Suppose we let $Z=X+Y$. We wish to find $Pr[Z=n]$. Note that we can write $Pr[Z=n]=\sum _{y=0}^{n}Pr[(X=n-y)\cap (Y=y)]=\sum _{y=0}^{n}Pr[X=n-y]Pr[Y=y].$

Now you know what each of these probabilities is, so write them out, and calculate the sum.

Kade Reese

Beginner2022-07-20Added 3 answers

Step 1

For the shifted geometric distribution, of X failures before success on trial $X+1$, $X\sim \mathcal{S}\mathcal{G}\mathcal{e}\mathcal{o}(p)\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}Pr(X=x)=p(1-p{)}^{x}$

$\begin{array}{rl}\text{Given:}& X\mathrm{\perp}Y,X\sim \mathcal{S}\mathcal{G}\mathcal{e}\mathcal{o}(p),Y\sim \mathcal{S}\mathcal{G}\mathcal{e}\mathcal{o}(p)\\ Pr(X+Y=n)& =Pr(\bigcup _{x=0}^{n}(X=x\cap Y=n-x))& \mathrm{\forall}n\in \{\mathrm{0..}\mathrm{\infty}\}\\ \text{}& =\sum _{x=0}^{n}Pr(X=x)Pr(Y=n-x)\\ \text{}& =\sum _{x=0}^{n}p(1-p{)}^{x}\cdot p(1-p{)}^{n-x}\\ \text{}& =(n+1){p}^{2}(1-p{)}^{n}\\ \text{}& \text{}\\ Pr(X=k\mid X+Y=n)& ={\displaystyle \frac{Pr(X=k\cap X+Y=n)}{Pr(X+Y=n)}}& \mathrm{\forall}n\in \{\mathrm{0..}\mathrm{\infty}\},k\in \{\mathrm{0..}n\}\\ \text{}& =\text{et cetera}\end{array}$

Step 2

For the geometric distribution of $X-1$ failures before success on trial X,

$X\sim \mathcal{G}\mathcal{e}\mathcal{o}(p)\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}Pr(X=x)=p(1-p{)}^{x-1}$

$\begin{array}{rl}\text{Given:}& X\mathrm{\perp}Y,X\sim \mathcal{G}\mathcal{e}\mathcal{o}(p),Y\sim \mathcal{G}\mathcal{e}\mathcal{o}(p)\\ Pr(X+Y=n)& =Pr(\bigcup _{x=1}^{n-1}(X=x\cap Y=n-x))& \mathrm{\forall}n\in \{\mathrm{2..}\mathrm{\infty}\}\\ \text{}& =\sum _{x=1}^{n-1}Pr(X=x)Pr(Y=n-x)\\ \text{}& =\sum _{x=1}^{n-1}p(1-p{)}^{x-1}\cdot p(1-p{)}^{n-x-1}\\ \text{}& =(n-1){p}^{2}(1-p{)}^{n-2}\\ \text{}& \text{}\\ Pr(X=k\mid X+Y=n)& ={\displaystyle \frac{Pr(X=k\cap X+Y=n)}{Pr(X+Y=n)}}& \mathrm{\forall}n\in \{\mathrm{2..}\mathrm{\infty}\},k\in \{\mathrm{1..}n-1\}\\ \text{}& =\text{et cetera}\end{array}$

For the shifted geometric distribution, of X failures before success on trial $X+1$, $X\sim \mathcal{S}\mathcal{G}\mathcal{e}\mathcal{o}(p)\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}Pr(X=x)=p(1-p{)}^{x}$

$\begin{array}{rl}\text{Given:}& X\mathrm{\perp}Y,X\sim \mathcal{S}\mathcal{G}\mathcal{e}\mathcal{o}(p),Y\sim \mathcal{S}\mathcal{G}\mathcal{e}\mathcal{o}(p)\\ Pr(X+Y=n)& =Pr(\bigcup _{x=0}^{n}(X=x\cap Y=n-x))& \mathrm{\forall}n\in \{\mathrm{0..}\mathrm{\infty}\}\\ \text{}& =\sum _{x=0}^{n}Pr(X=x)Pr(Y=n-x)\\ \text{}& =\sum _{x=0}^{n}p(1-p{)}^{x}\cdot p(1-p{)}^{n-x}\\ \text{}& =(n+1){p}^{2}(1-p{)}^{n}\\ \text{}& \text{}\\ Pr(X=k\mid X+Y=n)& ={\displaystyle \frac{Pr(X=k\cap X+Y=n)}{Pr(X+Y=n)}}& \mathrm{\forall}n\in \{\mathrm{0..}\mathrm{\infty}\},k\in \{\mathrm{0..}n\}\\ \text{}& =\text{et cetera}\end{array}$

Step 2

For the geometric distribution of $X-1$ failures before success on trial X,

$X\sim \mathcal{G}\mathcal{e}\mathcal{o}(p)\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}Pr(X=x)=p(1-p{)}^{x-1}$

$\begin{array}{rl}\text{Given:}& X\mathrm{\perp}Y,X\sim \mathcal{G}\mathcal{e}\mathcal{o}(p),Y\sim \mathcal{G}\mathcal{e}\mathcal{o}(p)\\ Pr(X+Y=n)& =Pr(\bigcup _{x=1}^{n-1}(X=x\cap Y=n-x))& \mathrm{\forall}n\in \{\mathrm{2..}\mathrm{\infty}\}\\ \text{}& =\sum _{x=1}^{n-1}Pr(X=x)Pr(Y=n-x)\\ \text{}& =\sum _{x=1}^{n-1}p(1-p{)}^{x-1}\cdot p(1-p{)}^{n-x-1}\\ \text{}& =(n-1){p}^{2}(1-p{)}^{n-2}\\ \text{}& \text{}\\ Pr(X=k\mid X+Y=n)& ={\displaystyle \frac{Pr(X=k\cap X+Y=n)}{Pr(X+Y=n)}}& \mathrm{\forall}n\in \{\mathrm{2..}\mathrm{\infty}\},k\in \{\mathrm{1..}n-1\}\\ \text{}& =\text{et cetera}\end{array}$

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