Mixing
Joint Distribution, Geometric Distribution
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Let X and Y be IID Geom(p) (independent and identically distributed geometric probability) and $N=X+Y$. Find the joint PMF (probability mass function) of $X,Y,N$.
I know the solution to this but I could not come up with it myself, and even after looking the solution up im not sure my reasoning is right.....
So the solution is
$$f(x,y,n)=f(N=x+y |X=x, Y=y) =1$$
therefore
$$f(x,y,n)=f(x,y)= pq^x *pq^y = p^2 * q^n$$
My reasoning for this is that if I Conidtion on x, y... they are true so if x and y are true than $n = x+y$ has to be true with $p=1$.
I know my explanation may sound stupid, i just started with probability after a i haven’t done any math for years.
I've also tried to reason mathematically but didn’t come very far...
$$
f_x(x)=pq^x-1\
f_y(y)=pq^y-1
$$
$f_n(n=X+Y)$ should be neg. Binomial $Nbin(2,p)$?
$binomn+11p^2q^n$
That is not very conclusive, and in particular, why is there a $q^x$ and not $x-1$?
There is so much i don’t get and unfortunately my books don’t help,
So many thanks for taking the time and reading this
negative-binomial
edited Jul 23 at 20:45
bjcolby15
8071716
asked Jul 23 at 19:44
Lillys
84
add a comment |Â
up vote
1
down vote
favorite
Let X and Y be IID Geom(p) (independent and identically distributed geometric probability) and $N=X+Y$. Find the joint PMF (probability mass function) of $X,Y,N$.
I know the solution to this but I could not come up with it myself, and even after looking the solution up im not sure my reasoning is right.....
So the solution is
$$f(x,y,n)=f(N=x+y |X=x, Y=y) =1$$
therefore
$$f(x,y,n)=f(x,y)= pq^x *pq^y = p^2 * q^n$$
My reasoning for this is that if I Conidtion on x, y... they are true so if x and y are true than $n = x+y$ has to be true with $p=1$.
I know my explanation may sound stupid, i just started with probability after a i haven’t done any math for years.
I've also tried to reason mathematically but didn’t come very far...
$$
f_x(x)=pq^x-1\
f_y(y)=pq^y-1
$$
$f_n(n=X+Y)$ should be neg. Binomial $Nbin(2,p)$?
$binomn+11p^2q^n$
That is not very conclusive, and in particular, why is there a $q^x$ and not $x-1$?
There is so much i don’t get and unfortunately my books don’t help,
So many thanks for taking the time and reading this
negative-binomial
edited Jul 23 at 20:45
bjcolby15
8071716
asked Jul 23 at 19:44
Lillys
84
They are independent so can I just take one side of the ‚conditioning‘? sorry I’m missing English vocabulary....
– Lillys
Jul 23 at 20:10
I looked up the acronyms for other readers to chime in...
– bjcolby15
Jul 23 at 20:49
1
Thanks, I will remember to write them out for my next question
– Lillys
Jul 23 at 21:07
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let X and Y be IID Geom(p) (independent and identically distributed geometric probability) and $N=X+Y$. Find the joint PMF (probability mass function) of $X,Y,N$.
I know the solution to this but I could not come up with it myself, and even after looking the solution up im not sure my reasoning is right.....
So the solution is
$$f(x,y,n)=f(N=x+y |X=x, Y=y) =1$$
therefore
$$f(x,y,n)=f(x,y)= pq^x *pq^y = p^2 * q^n$$
My reasoning for this is that if I Conidtion on x, y... they are true so if x and y are true than $n = x+y$ has to be true with $p=1$.
I know my explanation may sound stupid, i just started with probability after a i haven’t done any math for years.
I've also tried to reason mathematically but didn’t come very far...
$$
f_x(x)=pq^x-1\
f_y(y)=pq^y-1
$$
$f_n(n=X+Y)$ should be neg. Binomial $Nbin(2,p)$?
$binomn+11p^2q^n$
That is not very conclusive, and in particular, why is there a $q^x$ and not $x-1$?
There is so much i don’t get and unfortunately my books don’t help,
So many thanks for taking the time and reading this
negative-binomial
edited Jul 23 at 20:45
bjcolby15
8071716
asked Jul 23 at 19:44
Lillys
84
Let X and Y be IID Geom(p) (independent and identically distributed geometric probability) and $N=X+Y$. Find the joint PMF (probability mass function) of $X,Y,N$.
I know the solution to this but I could not come up with it myself, and even after looking the solution up im not sure my reasoning is right.....
So the solution is
$$f(x,y,n)=f(N=x+y |X=x, Y=y) =1$$
therefore
$$f(x,y,n)=f(x,y)= pq^x *pq^y = p^2 * q^n$$
My reasoning for this is that if I Conidtion on x, y... they are true so if x and y are true than $n = x+y$ has to be true with $p=1$.
I know my explanation may sound stupid, i just started with probability after a i haven’t done any math for years.
I've also tried to reason mathematically but didn’t come very far...
$$
f_x(x)=pq^x-1\
f_y(y)=pq^y-1
$$
$f_n(n=X+Y)$ should be neg. Binomial $Nbin(2,p)$?
$binomn+11p^2q^n$
That is not very conclusive, and in particular, why is there a $q^x$ and not $x-1$?
There is so much i don’t get and unfortunately my books don’t help,
So many thanks for taking the time and reading this
negative-binomial
edited Jul 23 at 20:45
bjcolby15
8071716
asked Jul 23 at 19:44
Lillys
84
edited Jul 23 at 20:45
bjcolby15
8071716
edited Jul 23 at 20:45
bjcolby15
8071716
edited Jul 23 at 20:45
bjcolby15
8071716
8071716
asked Jul 23 at 19:44
Lillys
84
asked Jul 23 at 19:44
Lillys
84
asked Jul 23 at 19:44
Lillys
84
84
They are independent so can I just take one side of the ‚conditioning‘? sorry I’m missing English vocabulary....
– Lillys
Jul 23 at 20:10
I looked up the acronyms for other readers to chime in...
– bjcolby15
Jul 23 at 20:49
1
Thanks, I will remember to write them out for my next question
– Lillys
Jul 23 at 21:07
add a comment |Â
They are independent so can I just take one side of the ‚conditioning‘? sorry I’m missing English vocabulary....
– Lillys
Jul 23 at 20:10
I looked up the acronyms for other readers to chime in...
– bjcolby15
Jul 23 at 20:49
1
Thanks, I will remember to write them out for my next question
– Lillys
Jul 23 at 21:07
They are independent so can I just take one side of the ‚conditioning‘? sorry I’m missing English vocabulary....
– Lillys
Jul 23 at 20:10
They are independent so can I just take one side of the ‚conditioning‘? sorry I’m missing English vocabulary....
– Lillys
Jul 23 at 20:10
I looked up the acronyms for other readers to chime in...
– bjcolby15
Jul 23 at 20:49
I looked up the acronyms for other readers to chime in...
– bjcolby15
Jul 23 at 20:49
1
1
Thanks, I will remember to write them out for my next question
– Lillys
Jul 23 at 21:07
Thanks, I will remember to write them out for my next question
– Lillys
Jul 23 at 21:07
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
1) Why is that $q^x$ but not $q^x-1$?
You need to look at the definition of the book that you use.
Some books define $Xsim Geom(p)$ as the total number of steps include the success. While some books define as the number of failures only (do not count the success).
In this solution, it seems your book uses the later definition. Thus the total steps will be x failures. Therefore:
$f_X=pq^x$ (x failures + 1 success)
$f_Y=pq^y$ (y failures + 1 success)
2)
You mistaken the probability of joint distribution of $(X,Y)$ with the distribution of N, the total steps.
The joint distribution of $(X,Y)$ tells the probability of one specific event $(X,Y)$ that satisfies $X + Y = N$:
$f_X,Y(x,y,n) = P(X=x,Y=n-x) = f_X(x)f_Y(n-x) $
$f_X,Y(x,y,n) = pq^xpq^n-x = p^2q^n$
While the distribution of N tells the probability of all events $(X,Y)$ that satisfy $X+Y=N$:
$f_N(n) = P(X+Y=N) = sum_x=0^nP(X=x)P(Y=n-x) = sum_x:0^n pq^xpq^n-x = (n+1)p^2q^n$
3) Another solution is to use Moment Generating Function. You can look it up.
answered Jul 24 at 22:27
tdluong
1315
add a comment |Â
up vote
0
down vote
Suppose that $X,Y sim Geometric(p) $
The joint pdf of two random variables is
$$f_X,Y(x,y) = f_X(x) f_Y(y) $$
Now the probability mass function the geometric random variable is given by
$$ p(k) = P(X=k) = (1-p)^k-1 p $$
then we have
$$ f_X,Y(x,y) = ((1-p)^x-1p)(1-p)^y-1p $$
$$ f_X,Y(x,y) = (1-p)^x+y-2p^2 $$
Then we have
$$N= X+Y sim NegBin(2,p) $$
The probability mass function for the negative binomial given by
$$ Xsim NegBin(r;p) $$
$$f(k;r,p) = Pr(X=k) = binomk+r-1k p^k(1-p)^r $$
In the geometric distribution we had our mass function
$$ Pr(X=k) = (1-p)^k-1p $$
then our number of failures are $$x+y-2 $$
$$ N = binomx+y-2+2-1x+y-2p^x+y-2(1-p)^2 $$
$$ N = binomx+y-1x+y-2p^x+y-2(1-p)^2 $$
Then we have
$$f_N,XY(n,xy) = f_N(n)f_X,Y(x,y) = binomx+y-1x+y-2 p^x+y-2(1-p)^2 (1-p)^x+y-2p^2 $$
Does this work?
answered Jul 23 at 22:38
RHowe
1,010815
Thanks for your answer, I’m sorry for answering so late, I’m moving at the moment so there is not much time left for studying, anyhow many thanks, just one or two questions. I have problems using the formula for the neg. Binomial... k=N=x+y... r would be two bc we have to Random variables.?. But why is the lower part of the Binomial X+y - 2?..degrees of freedom with the two random variables. I’m sorry if my questions sound stupid, I only start studying statistics in October, but I have some time now and so I’m studying ahead of course.....
– Lillys
Jul 25 at 20:14
I'm not sure if it's right..I had a class in it in the fall but your joint function gives the number of successes as $k=N=x+y-2$ the mass function in my book says the number of failures is $k-1$ failures so $p(k) = q^k-1p $ . You even have -1 .. if you multiple the two mass functions then $ q^(x-1) q^(y-1) = q^x+y-2)$
– RHowe
Jul 25 at 20:58
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
1) Why is that $q^x$ but not $q^x-1$?
You need to look at the definition of the book that you use.
Some books define $Xsim Geom(p)$ as the total number of steps include the success. While some books define as the number of failures only (do not count the success).
In this solution, it seems your book uses the later definition. Thus the total steps will be x failures. Therefore:
$f_X=pq^x$ (x failures + 1 success)
$f_Y=pq^y$ (y failures + 1 success)
2)
You mistaken the probability of joint distribution of $(X,Y)$ with the distribution of N, the total steps.
The joint distribution of $(X,Y)$ tells the probability of one specific event $(X,Y)$ that satisfies $X + Y = N$:
$f_X,Y(x,y,n) = P(X=x,Y=n-x) = f_X(x)f_Y(n-x) $
$f_X,Y(x,y,n) = pq^xpq^n-x = p^2q^n$
While the distribution of N tells the probability of all events $(X,Y)$ that satisfy $X+Y=N$:
$f_N(n) = P(X+Y=N) = sum_x=0^nP(X=x)P(Y=n-x) = sum_x:0^n pq^xpq^n-x = (n+1)p^2q^n$
3) Another solution is to use Moment Generating Function. You can look it up.
answered Jul 24 at 22:27
tdluong
1315
add a comment |Â
up vote
0
down vote
1) Why is that $q^x$ but not $q^x-1$?
You need to look at the definition of the book that you use.
Some books define $Xsim Geom(p)$ as the total number of steps include the success. While some books define as the number of failures only (do not count the success).
In this solution, it seems your book uses the later definition. Thus the total steps will be x failures. Therefore:
$f_X=pq^x$ (x failures + 1 success)
$f_Y=pq^y$ (y failures + 1 success)
2)
You mistaken the probability of joint distribution of $(X,Y)$ with the distribution of N, the total steps.
The joint distribution of $(X,Y)$ tells the probability of one specific event $(X,Y)$ that satisfies $X + Y = N$:
$f_X,Y(x,y,n) = P(X=x,Y=n-x) = f_X(x)f_Y(n-x) $
$f_X,Y(x,y,n) = pq^xpq^n-x = p^2q^n$
While the distribution of N tells the probability of all events $(X,Y)$ that satisfy $X+Y=N$:
$f_N(n) = P(X+Y=N) = sum_x=0^nP(X=x)P(Y=n-x) = sum_x:0^n pq^xpq^n-x = (n+1)p^2q^n$
3) Another solution is to use Moment Generating Function. You can look it up.
answered Jul 24 at 22:27
tdluong
1315
add a comment |Â
up vote
0
down vote
up vote
0
down vote
1) Why is that $q^x$ but not $q^x-1$?
You need to look at the definition of the book that you use.
Some books define $Xsim Geom(p)$ as the total number of steps include the success. While some books define as the number of failures only (do not count the success).
In this solution, it seems your book uses the later definition. Thus the total steps will be x failures. Therefore:
$f_X=pq^x$ (x failures + 1 success)
$f_Y=pq^y$ (y failures + 1 success)
2)
You mistaken the probability of joint distribution of $(X,Y)$ with the distribution of N, the total steps.
The joint distribution of $(X,Y)$ tells the probability of one specific event $(X,Y)$ that satisfies $X + Y = N$:
$f_X,Y(x,y,n) = P(X=x,Y=n-x) = f_X(x)f_Y(n-x) $
$f_X,Y(x,y,n) = pq^xpq^n-x = p^2q^n$
While the distribution of N tells the probability of all events $(X,Y)$ that satisfy $X+Y=N$:
$f_N(n) = P(X+Y=N) = sum_x=0^nP(X=x)P(Y=n-x) = sum_x:0^n pq^xpq^n-x = (n+1)p^2q^n$
3) Another solution is to use Moment Generating Function. You can look it up.
answered Jul 24 at 22:27
tdluong
1315
1) Why is that $q^x$ but not $q^x-1$?
You need to look at the definition of the book that you use.
Some books define $Xsim Geom(p)$ as the total number of steps include the success. While some books define as the number of failures only (do not count the success).
In this solution, it seems your book uses the later definition. Thus the total steps will be x failures. Therefore:
$f_X=pq^x$ (x failures + 1 success)
$f_Y=pq^y$ (y failures + 1 success)
2)
You mistaken the probability of joint distribution of $(X,Y)$ with the distribution of N, the total steps.
The joint distribution of $(X,Y)$ tells the probability of one specific event $(X,Y)$ that satisfies $X + Y = N$:
$f_X,Y(x,y,n) = P(X=x,Y=n-x) = f_X(x)f_Y(n-x) $
$f_X,Y(x,y,n) = pq^xpq^n-x = p^2q^n$
While the distribution of N tells the probability of all events $(X,Y)$ that satisfy $X+Y=N$:
$f_N(n) = P(X+Y=N) = sum_x=0^nP(X=x)P(Y=n-x) = sum_x:0^n pq^xpq^n-x = (n+1)p^2q^n$
3) Another solution is to use Moment Generating Function. You can look it up.
answered Jul 24 at 22:27
tdluong
1315
edited Jul 25 at 8:24
answered Jul 24 at 22:27
tdluong
1315
answered Jul 24 at 22:27
tdluong
1315
answered Jul 24 at 22:27
tdluong
1315
1315
add a comment |Â
add a comment |Â
up vote
0
down vote
Suppose that $X,Y sim Geometric(p) $
The joint pdf of two random variables is
$$f_X,Y(x,y) = f_X(x) f_Y(y) $$
Now the probability mass function the geometric random variable is given by
$$ p(k) = P(X=k) = (1-p)^k-1 p $$
then we have
$$ f_X,Y(x,y) = ((1-p)^x-1p)(1-p)^y-1p $$
$$ f_X,Y(x,y) = (1-p)^x+y-2p^2 $$
Then we have
$$N= X+Y sim NegBin(2,p) $$
The probability mass function for the negative binomial given by
$$ Xsim NegBin(r;p) $$
$$f(k;r,p) = Pr(X=k) = binomk+r-1k p^k(1-p)^r $$
In the geometric distribution we had our mass function
$$ Pr(X=k) = (1-p)^k-1p $$
then our number of failures are $$x+y-2 $$
$$ N = binomx+y-2+2-1x+y-2p^x+y-2(1-p)^2 $$
$$ N = binomx+y-1x+y-2p^x+y-2(1-p)^2 $$
Then we have
$$f_N,XY(n,xy) = f_N(n)f_X,Y(x,y) = binomx+y-1x+y-2 p^x+y-2(1-p)^2 (1-p)^x+y-2p^2 $$
Does this work?
answered Jul 23 at 22:38
RHowe
1,010815
Thanks for your answer, I’m sorry for answering so late, I’m moving at the moment so there is not much time left for studying, anyhow many thanks, just one or two questions. I have problems using the formula for the neg. Binomial... k=N=x+y... r would be two bc we have to Random variables.?. But why is the lower part of the Binomial X+y - 2?..degrees of freedom with the two random variables. I’m sorry if my questions sound stupid, I only start studying statistics in October, but I have some time now and so I’m studying ahead of course.....
– Lillys
Jul 25 at 20:14
I'm not sure if it's right..I had a class in it in the fall but your joint function gives the number of successes as $k=N=x+y-2$ the mass function in my book says the number of failures is $k-1$ failures so $p(k) = q^k-1p $ . You even have -1 .. if you multiple the two mass functions then $ q^(x-1) q^(y-1) = q^x+y-2)$
– RHowe
Jul 25 at 20:58
add a comment |Â
up vote
0
down vote
Suppose that $X,Y sim Geometric(p) $
The joint pdf of two random variables is
$$f_X,Y(x,y) = f_X(x) f_Y(y) $$
Now the probability mass function the geometric random variable is given by
$$ p(k) = P(X=k) = (1-p)^k-1 p $$
then we have
$$ f_X,Y(x,y) = ((1-p)^x-1p)(1-p)^y-1p $$
$$ f_X,Y(x,y) = (1-p)^x+y-2p^2 $$
Then we have
$$N= X+Y sim NegBin(2,p) $$
The probability mass function for the negative binomial given by
$$ Xsim NegBin(r;p) $$
$$f(k;r,p) = Pr(X=k) = binomk+r-1k p^k(1-p)^r $$
In the geometric distribution we had our mass function
$$ Pr(X=k) = (1-p)^k-1p $$
then our number of failures are $$x+y-2 $$
$$ N = binomx+y-2+2-1x+y-2p^x+y-2(1-p)^2 $$
$$ N = binomx+y-1x+y-2p^x+y-2(1-p)^2 $$
Then we have
$$f_N,XY(n,xy) = f_N(n)f_X,Y(x,y) = binomx+y-1x+y-2 p^x+y-2(1-p)^2 (1-p)^x+y-2p^2 $$
Does this work?
answered Jul 23 at 22:38
RHowe
1,010815
Thanks for your answer, I’m sorry for answering so late, I’m moving at the moment so there is not much time left for studying, anyhow many thanks, just one or two questions. I have problems using the formula for the neg. Binomial... k=N=x+y... r would be two bc we have to Random variables.?. But why is the lower part of the Binomial X+y - 2?..degrees of freedom with the two random variables. I’m sorry if my questions sound stupid, I only start studying statistics in October, but I have some time now and so I’m studying ahead of course.....
– Lillys
Jul 25 at 20:14
I'm not sure if it's right..I had a class in it in the fall but your joint function gives the number of successes as $k=N=x+y-2$ the mass function in my book says the number of failures is $k-1$ failures so $p(k) = q^k-1p $ . You even have -1 .. if you multiple the two mass functions then $ q^(x-1) q^(y-1) = q^x+y-2)$
– RHowe
Jul 25 at 20:58
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Suppose that $X,Y sim Geometric(p) $
The joint pdf of two random variables is
$$f_X,Y(x,y) = f_X(x) f_Y(y) $$
Now the probability mass function the geometric random variable is given by
$$ p(k) = P(X=k) = (1-p)^k-1 p $$
then we have
$$ f_X,Y(x,y) = ((1-p)^x-1p)(1-p)^y-1p $$
$$ f_X,Y(x,y) = (1-p)^x+y-2p^2 $$
Then we have
$$N= X+Y sim NegBin(2,p) $$
The probability mass function for the negative binomial given by
$$ Xsim NegBin(r;p) $$
$$f(k;r,p) = Pr(X=k) = binomk+r-1k p^k(1-p)^r $$
In the geometric distribution we had our mass function
$$ Pr(X=k) = (1-p)^k-1p $$
then our number of failures are $$x+y-2 $$
$$ N = binomx+y-2+2-1x+y-2p^x+y-2(1-p)^2 $$
$$ N = binomx+y-1x+y-2p^x+y-2(1-p)^2 $$
Then we have
$$f_N,XY(n,xy) = f_N(n)f_X,Y(x,y) = binomx+y-1x+y-2 p^x+y-2(1-p)^2 (1-p)^x+y-2p^2 $$
Does this work?
answered Jul 23 at 22:38
RHowe
1,010815
Suppose that $X,Y sim Geometric(p) $
The joint pdf of two random variables is
$$f_X,Y(x,y) = f_X(x) f_Y(y) $$
Now the probability mass function the geometric random variable is given by
$$ p(k) = P(X=k) = (1-p)^k-1 p $$
then we have
$$ f_X,Y(x,y) = ((1-p)^x-1p)(1-p)^y-1p $$
$$ f_X,Y(x,y) = (1-p)^x+y-2p^2 $$
Then we have
$$N= X+Y sim NegBin(2,p) $$
The probability mass function for the negative binomial given by
$$ Xsim NegBin(r;p) $$
$$f(k;r,p) = Pr(X=k) = binomk+r-1k p^k(1-p)^r $$
In the geometric distribution we had our mass function
$$ Pr(X=k) = (1-p)^k-1p $$
then our number of failures are $$x+y-2 $$
$$ N = binomx+y-2+2-1x+y-2p^x+y-2(1-p)^2 $$
$$ N = binomx+y-1x+y-2p^x+y-2(1-p)^2 $$
Then we have
$$f_N,XY(n,xy) = f_N(n)f_X,Y(x,y) = binomx+y-1x+y-2 p^x+y-2(1-p)^2 (1-p)^x+y-2p^2 $$
Does this work?
answered Jul 23 at 22:38
RHowe
1,010815
edited Jul 25 at 21:01
answered Jul 23 at 22:38
RHowe
1,010815
answered Jul 23 at 22:38
RHowe
1,010815
answered Jul 23 at 22:38
RHowe
1,010815
1,010815
Thanks for your answer, I’m sorry for answering so late, I’m moving at the moment so there is not much time left for studying, anyhow many thanks, just one or two questions. I have problems using the formula for the neg. Binomial... k=N=x+y... r would be two bc we have to Random variables.?. But why is the lower part of the Binomial X+y - 2?..degrees of freedom with the two random variables. I’m sorry if my questions sound stupid, I only start studying statistics in October, but I have some time now and so I’m studying ahead of course.....
– Lillys
Jul 25 at 20:14
I'm not sure if it's right..I had a class in it in the fall but your joint function gives the number of successes as $k=N=x+y-2$ the mass function in my book says the number of failures is $k-1$ failures so $p(k) = q^k-1p $ . You even have -1 .. if you multiple the two mass functions then $ q^(x-1) q^(y-1) = q^x+y-2)$
– RHowe
Jul 25 at 20:58
add a comment |Â
Thanks for your answer, I’m sorry for answering so late, I’m moving at the moment so there is not much time left for studying, anyhow many thanks, just one or two questions. I have problems using the formula for the neg. Binomial... k=N=x+y... r would be two bc we have to Random variables.?. But why is the lower part of the Binomial X+y - 2?..degrees of freedom with the two random variables. I’m sorry if my questions sound stupid, I only start studying statistics in October, but I have some time now and so I’m studying ahead of course.....
– Lillys
Jul 25 at 20:14
I'm not sure if it's right..I had a class in it in the fall but your joint function gives the number of successes as $k=N=x+y-2$ the mass function in my book says the number of failures is $k-1$ failures so $p(k) = q^k-1p $ . You even have -1 .. if you multiple the two mass functions then $ q^(x-1) q^(y-1) = q^x+y-2)$
– RHowe
Jul 25 at 20:58
Thanks for your answer, I’m sorry for answering so late, I’m moving at the moment so there is not much time left for studying, anyhow many thanks, just one or two questions. I have problems using the formula for the neg. Binomial... k=N=x+y... r would be two bc we have to Random variables.?. But why is the lower part of the Binomial X+y - 2?..degrees of freedom with the two random variables. I’m sorry if my questions sound stupid, I only start studying statistics in October, but I have some time now and so I’m studying ahead of course.....
– Lillys
Jul 25 at 20:14
Thanks for your answer, I’m sorry for answering so late, I’m moving at the moment so there is not much time left for studying, anyhow many thanks, just one or two questions. I have problems using the formula for the neg. Binomial... k=N=x+y... r would be two bc we have to Random variables.?. But why is the lower part of the Binomial X+y - 2?..degrees of freedom with the two random variables. I’m sorry if my questions sound stupid, I only start studying statistics in October, but I have some time now and so I’m studying ahead of course.....
– Lillys
Jul 25 at 20:14
I'm not sure if it's right..I had a class in it in the fall but your joint function gives the number of successes as $k=N=x+y-2$ the mass function in my book says the number of failures is $k-1$ failures so $p(k) = q^k-1p $ . You even have -1 .. if you multiple the two mass functions then $ q^(x-1) q^(y-1) = q^x+y-2)$
– RHowe
Jul 25 at 20:58
I'm not sure if it's right..I had a class in it in the fall but your joint function gives the number of successes as $k=N=x+y-2$ the mass function in my book says the number of failures is $k-1$ failures so $p(k) = q^k-1p $ . You even have -1 .. if you multiple the two mass functions then $ q^(x-1) q^(y-1) = q^x+y-2)$
– RHowe
Jul 25 at 20:58
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2860723%2fjoint-distribution-geometric-distribution%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
They are independent so can I just take one side of the ‚conditioning‘? sorry I’m missing English vocabulary....
– Lillys
Jul 23 at 20:10
I looked up the acronyms for other readers to chime in...
– bjcolby15
Jul 23 at 20:49
1
Thanks, I will remember to write them out for my next question
– Lillys
Jul 23 at 21:07