Sum of two random variables ( negative binomial distribution )
$begingroup$
Let $X,Y$ be two independent negative binomial distributed random variables.
$X$ ~ $NB(r,p)$ and $Y$ ~ $NB(s,p)$
Show that:
$ X+Y $ ~ $ NB(r+s,p) $.
Remark: So where I'm stucked?
I failed to show that $ sum_{j=0}^{k} binom{j+r-1}{j} cdot binom{k-j+s-1}{k-j} = binom{k+r+s-1}{k}$. If I have this identity I can solve this exercise. First I thought that this is the vandermonde identity, but it isn't. So how can I show this identity? I know that I can solve this exercise by using the fact that a negative binomial distributed RV is a sum of geometric distributed RV, but i want to show it with my attempt.
Thank you for your help.
probability-theory probability-distributions random-variables negative-binomial
asked Dec 2 '17 at 12:06
RukiaKuchikiRukiaKuchiki
337211
$endgroup$
add a comment |
$begingroup$
Let $X,Y$ be two independent negative binomial distributed random variables.
$X$ ~ $NB(r,p)$ and $Y$ ~ $NB(s,p)$
Show that:
$ X+Y $ ~ $ NB(r+s,p) $.
Remark: So where I'm stucked?
I failed to show that $ sum_{j=0}^{k} binom{j+r-1}{j} cdot binom{k-j+s-1}{k-j} = binom{k+r+s-1}{k}$. If I have this identity I can solve this exercise. First I thought that this is the vandermonde identity, but it isn't. So how can I show this identity? I know that I can solve this exercise by using the fact that a negative binomial distributed RV is a sum of geometric distributed RV, but i want to show it with my attempt.
Thank you for your help.
probability-theory probability-distributions random-variables negative-binomial
asked Dec 2 '17 at 12:06
RukiaKuchikiRukiaKuchiki
337211
$endgroup$
add a comment |
$begingroup$
Let $X,Y$ be two independent negative binomial distributed random variables.
$X$ ~ $NB(r,p)$ and $Y$ ~ $NB(s,p)$
Show that:
$ X+Y $ ~ $ NB(r+s,p) $.
Remark: So where I'm stucked?
I failed to show that $ sum_{j=0}^{k} binom{j+r-1}{j} cdot binom{k-j+s-1}{k-j} = binom{k+r+s-1}{k}$. If I have this identity I can solve this exercise. First I thought that this is the vandermonde identity, but it isn't. So how can I show this identity? I know that I can solve this exercise by using the fact that a negative binomial distributed RV is a sum of geometric distributed RV, but i want to show it with my attempt.
Thank you for your help.
probability-theory probability-distributions random-variables negative-binomial
asked Dec 2 '17 at 12:06
RukiaKuchikiRukiaKuchiki
337211
$endgroup$
Let $X,Y$ be two independent negative binomial distributed random variables.
$X$ ~ $NB(r,p)$ and $Y$ ~ $NB(s,p)$
Show that:
$ X+Y $ ~ $ NB(r+s,p) $.
Remark: So where I'm stucked?
I failed to show that $ sum_{j=0}^{k} binom{j+r-1}{j} cdot binom{k-j+s-1}{k-j} = binom{k+r+s-1}{k}$. If I have this identity I can solve this exercise. First I thought that this is the vandermonde identity, but it isn't. So how can I show this identity? I know that I can solve this exercise by using the fact that a negative binomial distributed RV is a sum of geometric distributed RV, but i want to show it with my attempt.
Thank you for your help.
probability-theory probability-distributions random-variables negative-binomial
probability-theory probability-distributions random-variables negative-binomial
asked Dec 2 '17 at 12:06
RukiaKuchikiRukiaKuchiki
337211
asked Dec 2 '17 at 12:06
RukiaKuchikiRukiaKuchiki
337211
asked Dec 2 '17 at 12:06
RukiaKuchikiRukiaKuchiki
337211
asked Dec 2 '17 at 12:06
RukiaKuchikiRukiaKuchiki
337211
asked Dec 2 '17 at 12:06
RukiaKuchikiRukiaKuchiki
337211
337211
add a comment |
add a comment |
1 Answer
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$begingroup$
In general we have:$$sum_{i+j=k}binom{i}{r}binom{j}{s}=binom{k+1}{r+s+1}tag1$$
This under the convention that $binom{n}{k}=0$ if $knotin{0,1,dots,n}$.
For a combinatorial proof think of a row of $k+1$ balls.
Selecting $r+s+1$ of them can be done on $binom{k+1}{r+s+1}$ ways (RHS).
Doing so we first pick out a ball that cuts the row in a left row of length $igeq r$ balls and a right row of length $jgeq s$ balls. So this under the condition that $i+j=k$. Then from the left row we select $r$ balls and from the right row we select $s$ balls. This process reveals the LHS.
Your summation can be rewritten by:$$sum_{i+j=k}binom{j+r-1}{r-1}binom{i+s-1}{s-1}=sum_{i+j=k+r+s-2}binom{j}{r-1}binom{i}{s-1}$$
Then applying $(1)$ gives:$$=binom{k+r+s-1}{r+s-1}=binom{k+r+s-1}{k}$$
answered Dec 2 '17 at 12:43
drhabdrhab
101k545136
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
In general we have:$$sum_{i+j=k}binom{i}{r}binom{j}{s}=binom{k+1}{r+s+1}tag1$$
This under the convention that $binom{n}{k}=0$ if $knotin{0,1,dots,n}$.
For a combinatorial proof think of a row of $k+1$ balls.
Selecting $r+s+1$ of them can be done on $binom{k+1}{r+s+1}$ ways (RHS).
Doing so we first pick out a ball that cuts the row in a left row of length $igeq r$ balls and a right row of length $jgeq s$ balls. So this under the condition that $i+j=k$. Then from the left row we select $r$ balls and from the right row we select $s$ balls. This process reveals the LHS.
Your summation can be rewritten by:$$sum_{i+j=k}binom{j+r-1}{r-1}binom{i+s-1}{s-1}=sum_{i+j=k+r+s-2}binom{j}{r-1}binom{i}{s-1}$$
Then applying $(1)$ gives:$$=binom{k+r+s-1}{r+s-1}=binom{k+r+s-1}{k}$$
answered Dec 2 '17 at 12:43
drhabdrhab
101k545136
$endgroup$
add a comment |
$begingroup$
In general we have:$$sum_{i+j=k}binom{i}{r}binom{j}{s}=binom{k+1}{r+s+1}tag1$$
This under the convention that $binom{n}{k}=0$ if $knotin{0,1,dots,n}$.
For a combinatorial proof think of a row of $k+1$ balls.
Selecting $r+s+1$ of them can be done on $binom{k+1}{r+s+1}$ ways (RHS).
Doing so we first pick out a ball that cuts the row in a left row of length $igeq r$ balls and a right row of length $jgeq s$ balls. So this under the condition that $i+j=k$. Then from the left row we select $r$ balls and from the right row we select $s$ balls. This process reveals the LHS.
Your summation can be rewritten by:$$sum_{i+j=k}binom{j+r-1}{r-1}binom{i+s-1}{s-1}=sum_{i+j=k+r+s-2}binom{j}{r-1}binom{i}{s-1}$$
Then applying $(1)$ gives:$$=binom{k+r+s-1}{r+s-1}=binom{k+r+s-1}{k}$$
answered Dec 2 '17 at 12:43
drhabdrhab
101k545136
$endgroup$
add a comment |
$begingroup$
In general we have:$$sum_{i+j=k}binom{i}{r}binom{j}{s}=binom{k+1}{r+s+1}tag1$$
This under the convention that $binom{n}{k}=0$ if $knotin{0,1,dots,n}$.
For a combinatorial proof think of a row of $k+1$ balls.
Selecting $r+s+1$ of them can be done on $binom{k+1}{r+s+1}$ ways (RHS).
Doing so we first pick out a ball that cuts the row in a left row of length $igeq r$ balls and a right row of length $jgeq s$ balls. So this under the condition that $i+j=k$. Then from the left row we select $r$ balls and from the right row we select $s$ balls. This process reveals the LHS.
Your summation can be rewritten by:$$sum_{i+j=k}binom{j+r-1}{r-1}binom{i+s-1}{s-1}=sum_{i+j=k+r+s-2}binom{j}{r-1}binom{i}{s-1}$$
Then applying $(1)$ gives:$$=binom{k+r+s-1}{r+s-1}=binom{k+r+s-1}{k}$$
answered Dec 2 '17 at 12:43
drhabdrhab
101k545136
$endgroup$
In general we have:$$sum_{i+j=k}binom{i}{r}binom{j}{s}=binom{k+1}{r+s+1}tag1$$
This under the convention that $binom{n}{k}=0$ if $knotin{0,1,dots,n}$.
For a combinatorial proof think of a row of $k+1$ balls.
Selecting $r+s+1$ of them can be done on $binom{k+1}{r+s+1}$ ways (RHS).
Doing so we first pick out a ball that cuts the row in a left row of length $igeq r$ balls and a right row of length $jgeq s$ balls. So this under the condition that $i+j=k$. Then from the left row we select $r$ balls and from the right row we select $s$ balls. This process reveals the LHS.
Your summation can be rewritten by:$$sum_{i+j=k}binom{j+r-1}{r-1}binom{i+s-1}{s-1}=sum_{i+j=k+r+s-2}binom{j}{r-1}binom{i}{s-1}$$
Then applying $(1)$ gives:$$=binom{k+r+s-1}{r+s-1}=binom{k+r+s-1}{k}$$
answered Dec 2 '17 at 12:43
drhabdrhab
101k545136
edited Dec 9 '18 at 7:50
answered Dec 2 '17 at 12:43
drhabdrhab
101k545136
answered Dec 2 '17 at 12:43
drhabdrhab
101k545136
answered Dec 2 '17 at 12:43
drhabdrhab
101k545136
101k545136
add a comment |
add a comment |
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