Mixing
Relation between Poisson process and geometric distribution
$begingroup$
Suppose that red cars arrive at an intersection according to a Poisson process with rate parameter $r>0$ and blue cars arrive, independently of red cars, according to a Poisson process with rate parameter $b>0$. If $X$ is the number of blue cars that arrive between two successive red cars, show that $X$ has a geometric distribution.
Can anyone help me with this problem? I have tried to rephrase $X$ as: how many cars will arrive at the intersection before we see any red car.
probability poisson-process geometric-probability
edited Dec 10 '16 at 9:01
L.V.Rao
1,7491617
asked Dec 10 '16 at 2:30
Al GaussAl Gauss
63
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add a comment |
$begingroup$
Suppose that red cars arrive at an intersection according to a Poisson process with rate parameter $r>0$ and blue cars arrive, independently of red cars, according to a Poisson process with rate parameter $b>0$. If $X$ is the number of blue cars that arrive between two successive red cars, show that $X$ has a geometric distribution.
Can anyone help me with this problem? I have tried to rephrase $X$ as: how many cars will arrive at the intersection before we see any red car.
probability poisson-process geometric-probability
edited Dec 10 '16 at 9:01
L.V.Rao
1,7491617
asked Dec 10 '16 at 2:30
Al GaussAl Gauss
63
$endgroup$
$begingroup$
See my answer to this question
$endgroup$
– i707107
Dec 10 '16 at 3:18
$begingroup$
http://math.stackexchange.com/questions/262286/integration-help-from-poisson-process-question
$endgroup$
– Momo
Dec 10 '16 at 3:20
add a comment |
$begingroup$
Suppose that red cars arrive at an intersection according to a Poisson process with rate parameter $r>0$ and blue cars arrive, independently of red cars, according to a Poisson process with rate parameter $b>0$. If $X$ is the number of blue cars that arrive between two successive red cars, show that $X$ has a geometric distribution.
Can anyone help me with this problem? I have tried to rephrase $X$ as: how many cars will arrive at the intersection before we see any red car.
probability poisson-process geometric-probability
edited Dec 10 '16 at 9:01
L.V.Rao
1,7491617
asked Dec 10 '16 at 2:30
Al GaussAl Gauss
63
$endgroup$
Suppose that red cars arrive at an intersection according to a Poisson process with rate parameter $r>0$ and blue cars arrive, independently of red cars, according to a Poisson process with rate parameter $b>0$. If $X$ is the number of blue cars that arrive between two successive red cars, show that $X$ has a geometric distribution.
Can anyone help me with this problem? I have tried to rephrase $X$ as: how many cars will arrive at the intersection before we see any red car.
probability poisson-process geometric-probability
probability poisson-process geometric-probability
edited Dec 10 '16 at 9:01
L.V.Rao
1,7491617
asked Dec 10 '16 at 2:30
Al GaussAl Gauss
63
edited Dec 10 '16 at 9:01
L.V.Rao
1,7491617
asked Dec 10 '16 at 2:30
Al GaussAl Gauss
63
edited Dec 10 '16 at 9:01
L.V.Rao
1,7491617
edited Dec 10 '16 at 9:01
L.V.Rao
1,7491617
edited Dec 10 '16 at 9:01
L.V.Rao
1,7491617
1,7491617
asked Dec 10 '16 at 2:30
Al GaussAl Gauss
63
asked Dec 10 '16 at 2:30
Al GaussAl Gauss
63
asked Dec 10 '16 at 2:30
Al GaussAl Gauss
63
63
$begingroup$
See my answer to this question
$endgroup$
– i707107
Dec 10 '16 at 3:18
$begingroup$
http://math.stackexchange.com/questions/262286/integration-help-from-poisson-process-question
$endgroup$
– Momo
Dec 10 '16 at 3:20
add a comment |
$begingroup$
See my answer to this question
$endgroup$
– i707107
Dec 10 '16 at 3:18
$begingroup$
http://math.stackexchange.com/questions/262286/integration-help-from-poisson-process-question
$endgroup$
– Momo
Dec 10 '16 at 3:20
$begingroup$
See my answer to this question
$endgroup$
– i707107
Dec 10 '16 at 3:18
$begingroup$
See my answer to this question
$endgroup$
– i707107
Dec 10 '16 at 3:18
$begingroup$
http://math.stackexchange.com/questions/262286/integration-help-from-poisson-process-question
$endgroup$
– Momo
Dec 10 '16 at 3:20
$begingroup$
http://math.stackexchange.com/questions/262286/integration-help-from-poisson-process-question
$endgroup$
– Momo
Dec 10 '16 at 3:20
add a comment |
1 Answer
1
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oldest
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$begingroup$
Let the arrival process of red cars is according to the Poisson process, say, ${R(t),tge 0}$ with an intensity parameter $lambda_1>0$ and the arrival process of blue cars is according to the Poisson process, say, ${B(t),tge 0}$ with an intensity parameter $lambda_2>0$. Further, both the processes are independent of each other. We are interested in the number of arrivals of blue cars between two successive arrivals of red cars.
We know that, the inter-arrival times of a Poisson process with parameter $lambda$ are exponential with mean $1/lambda$. This means that, the pdf of inter arrival times between successive red cars is $$f(t)=lambda_1 e^{-lambda_1 t}.$$ Now, the probability distribution of $X$, the number of blue cars during an arbitrary interval (as determined by successive arrivals of red cars) is
begin{eqnarray*}
P{X=k}&=&int_{0}^{infty}dfrac{e^{-lambda_2 t}(lambda_{2}t)^{k}}{k!} f(t)dt\
&=&int_{0}^{infty}dfrac{e^{-lambda_2 t}(lambda_{2}t)^{k}}{k!} left(lambda_1 e^{-lambda_1 t}right) dt\
&=&dfrac{lambda_1 lambda_{2}^{k}}{k!}int_{0}^{infty}e^{-(lambda_1 +lambda_2)t}cdot t^{(k+1)-1}dt\
&=&dfrac{lambda_1 lambda_{2}^{k}}{k!} cdot dfrac{Gamma (k+1)}{(lambda_1 +lambda_2)^{k+1}}\
P{X=k}&=&left(dfrac{lambda_1}{lambda_1 +lambda_2}right)left(dfrac{lambda_2}{lambda_1 +lambda_2}right)^{k},quad k=0,1,2,cdots
end{eqnarray*}
which is a geometric distribution with parameter $dfrac{lambda_1}{lambda_1 +lambda_2}$.
answered Dec 10 '16 at 6:01
L.V.RaoL.V.Rao
1,7491617
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let the arrival process of red cars is according to the Poisson process, say, ${R(t),tge 0}$ with an intensity parameter $lambda_1>0$ and the arrival process of blue cars is according to the Poisson process, say, ${B(t),tge 0}$ with an intensity parameter $lambda_2>0$. Further, both the processes are independent of each other. We are interested in the number of arrivals of blue cars between two successive arrivals of red cars.
We know that, the inter-arrival times of a Poisson process with parameter $lambda$ are exponential with mean $1/lambda$. This means that, the pdf of inter arrival times between successive red cars is $$f(t)=lambda_1 e^{-lambda_1 t}.$$ Now, the probability distribution of $X$, the number of blue cars during an arbitrary interval (as determined by successive arrivals of red cars) is
begin{eqnarray*}
P{X=k}&=&int_{0}^{infty}dfrac{e^{-lambda_2 t}(lambda_{2}t)^{k}}{k!} f(t)dt\
&=&int_{0}^{infty}dfrac{e^{-lambda_2 t}(lambda_{2}t)^{k}}{k!} left(lambda_1 e^{-lambda_1 t}right) dt\
&=&dfrac{lambda_1 lambda_{2}^{k}}{k!}int_{0}^{infty}e^{-(lambda_1 +lambda_2)t}cdot t^{(k+1)-1}dt\
&=&dfrac{lambda_1 lambda_{2}^{k}}{k!} cdot dfrac{Gamma (k+1)}{(lambda_1 +lambda_2)^{k+1}}\
P{X=k}&=&left(dfrac{lambda_1}{lambda_1 +lambda_2}right)left(dfrac{lambda_2}{lambda_1 +lambda_2}right)^{k},quad k=0,1,2,cdots
end{eqnarray*}
which is a geometric distribution with parameter $dfrac{lambda_1}{lambda_1 +lambda_2}$.
answered Dec 10 '16 at 6:01
L.V.RaoL.V.Rao
1,7491617
$endgroup$
add a comment |
$begingroup$
Let the arrival process of red cars is according to the Poisson process, say, ${R(t),tge 0}$ with an intensity parameter $lambda_1>0$ and the arrival process of blue cars is according to the Poisson process, say, ${B(t),tge 0}$ with an intensity parameter $lambda_2>0$. Further, both the processes are independent of each other. We are interested in the number of arrivals of blue cars between two successive arrivals of red cars.
We know that, the inter-arrival times of a Poisson process with parameter $lambda$ are exponential with mean $1/lambda$. This means that, the pdf of inter arrival times between successive red cars is $$f(t)=lambda_1 e^{-lambda_1 t}.$$ Now, the probability distribution of $X$, the number of blue cars during an arbitrary interval (as determined by successive arrivals of red cars) is
begin{eqnarray*}
P{X=k}&=&int_{0}^{infty}dfrac{e^{-lambda_2 t}(lambda_{2}t)^{k}}{k!} f(t)dt\
&=&int_{0}^{infty}dfrac{e^{-lambda_2 t}(lambda_{2}t)^{k}}{k!} left(lambda_1 e^{-lambda_1 t}right) dt\
&=&dfrac{lambda_1 lambda_{2}^{k}}{k!}int_{0}^{infty}e^{-(lambda_1 +lambda_2)t}cdot t^{(k+1)-1}dt\
&=&dfrac{lambda_1 lambda_{2}^{k}}{k!} cdot dfrac{Gamma (k+1)}{(lambda_1 +lambda_2)^{k+1}}\
P{X=k}&=&left(dfrac{lambda_1}{lambda_1 +lambda_2}right)left(dfrac{lambda_2}{lambda_1 +lambda_2}right)^{k},quad k=0,1,2,cdots
end{eqnarray*}
which is a geometric distribution with parameter $dfrac{lambda_1}{lambda_1 +lambda_2}$.
answered Dec 10 '16 at 6:01
L.V.RaoL.V.Rao
1,7491617
$endgroup$
add a comment |
$begingroup$
Let the arrival process of red cars is according to the Poisson process, say, ${R(t),tge 0}$ with an intensity parameter $lambda_1>0$ and the arrival process of blue cars is according to the Poisson process, say, ${B(t),tge 0}$ with an intensity parameter $lambda_2>0$. Further, both the processes are independent of each other. We are interested in the number of arrivals of blue cars between two successive arrivals of red cars.
We know that, the inter-arrival times of a Poisson process with parameter $lambda$ are exponential with mean $1/lambda$. This means that, the pdf of inter arrival times between successive red cars is $$f(t)=lambda_1 e^{-lambda_1 t}.$$ Now, the probability distribution of $X$, the number of blue cars during an arbitrary interval (as determined by successive arrivals of red cars) is
begin{eqnarray*}
P{X=k}&=&int_{0}^{infty}dfrac{e^{-lambda_2 t}(lambda_{2}t)^{k}}{k!} f(t)dt\
&=&int_{0}^{infty}dfrac{e^{-lambda_2 t}(lambda_{2}t)^{k}}{k!} left(lambda_1 e^{-lambda_1 t}right) dt\
&=&dfrac{lambda_1 lambda_{2}^{k}}{k!}int_{0}^{infty}e^{-(lambda_1 +lambda_2)t}cdot t^{(k+1)-1}dt\
&=&dfrac{lambda_1 lambda_{2}^{k}}{k!} cdot dfrac{Gamma (k+1)}{(lambda_1 +lambda_2)^{k+1}}\
P{X=k}&=&left(dfrac{lambda_1}{lambda_1 +lambda_2}right)left(dfrac{lambda_2}{lambda_1 +lambda_2}right)^{k},quad k=0,1,2,cdots
end{eqnarray*}
which is a geometric distribution with parameter $dfrac{lambda_1}{lambda_1 +lambda_2}$.
answered Dec 10 '16 at 6:01
L.V.RaoL.V.Rao
1,7491617
$endgroup$
Let the arrival process of red cars is according to the Poisson process, say, ${R(t),tge 0}$ with an intensity parameter $lambda_1>0$ and the arrival process of blue cars is according to the Poisson process, say, ${B(t),tge 0}$ with an intensity parameter $lambda_2>0$. Further, both the processes are independent of each other. We are interested in the number of arrivals of blue cars between two successive arrivals of red cars.
We know that, the inter-arrival times of a Poisson process with parameter $lambda$ are exponential with mean $1/lambda$. This means that, the pdf of inter arrival times between successive red cars is $$f(t)=lambda_1 e^{-lambda_1 t}.$$ Now, the probability distribution of $X$, the number of blue cars during an arbitrary interval (as determined by successive arrivals of red cars) is
begin{eqnarray*}
P{X=k}&=&int_{0}^{infty}dfrac{e^{-lambda_2 t}(lambda_{2}t)^{k}}{k!} f(t)dt\
&=&int_{0}^{infty}dfrac{e^{-lambda_2 t}(lambda_{2}t)^{k}}{k!} left(lambda_1 e^{-lambda_1 t}right) dt\
&=&dfrac{lambda_1 lambda_{2}^{k}}{k!}int_{0}^{infty}e^{-(lambda_1 +lambda_2)t}cdot t^{(k+1)-1}dt\
&=&dfrac{lambda_1 lambda_{2}^{k}}{k!} cdot dfrac{Gamma (k+1)}{(lambda_1 +lambda_2)^{k+1}}\
P{X=k}&=&left(dfrac{lambda_1}{lambda_1 +lambda_2}right)left(dfrac{lambda_2}{lambda_1 +lambda_2}right)^{k},quad k=0,1,2,cdots
end{eqnarray*}
which is a geometric distribution with parameter $dfrac{lambda_1}{lambda_1 +lambda_2}$.
answered Dec 10 '16 at 6:01
L.V.RaoL.V.Rao
1,7491617
answered Dec 10 '16 at 6:01
L.V.RaoL.V.Rao
1,7491617
answered Dec 10 '16 at 6:01
L.V.RaoL.V.Rao
1,7491617
answered Dec 10 '16 at 6:01
L.V.RaoL.V.Rao
1,7491617
1,7491617
add a comment |
add a comment |
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$begingroup$
See my answer to this question
$endgroup$
– i707107
Dec 10 '16 at 3:18
$begingroup$
http://math.stackexchange.com/questions/262286/integration-help-from-poisson-process-question
$endgroup$
– Momo
Dec 10 '16 at 3:20