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Joint Distribution for Number of Arrivals
The Next CEO of Stack OverflowSplitting Poisson process formal proofThe Poisson DistributionWhat is the probability that a customer waits for lesser than 3 minutes?Poisson process - number of store purchases in a given timeSum of random variables and joint distributionPoisson process different type of eventsIdentifying a Poisson distributionFirst arrival timingsArrival Time in Poisson ProcessHow to solve this problem with Poisson distributionPoisson Process with Stationary Arrival Rate - Conditional Arrivals
$begingroup$
Suppose that the number of women who buy concert tickets follows a Poisson process with rate $30$ women per hour, and similarly the number of men who buy tickets is a Poisson distribution with rate $20$ per hour.
Regardless of gender customers buy
$1$ ticket with probability $1/2$
$2$ tickets with probability $2/5$
$3$ tickets with probabiltiy $1/10$
Letting $N_i$ be the number of customers that buy $i$ tickets in the first hour, find the joint distribution of $(N_1,N_2,N_3)$.
I know that since the number of females $N_F$ and $N_M$ are independent, that their superposition $N_F+N_M$ is a Poisson process with rate $30+20=50$, and also that the arrival times follow an ordered uniform distribution. But I don't what a joint distribution on the vector $(N_1,N_2,N_3)$ would even look like. Any help is appreciated.
Attempt.
Since $N_1+N_2+N_3=N(1)$ is a Poisson process with rate $50$, then
$$N_1sim Poisson(frac12cdot 50)$$
is a Poisson process, and similarly for $N_2$ and $N_3$,
$$N_2sim Poisson(frac25cdot 50)$$
$$N_3sim Poisson(frac110cdot 50)$$
So the joint distribution function is given by
$$P(N_1=i_1,N_2=i_2,N_3=i_3)=prod_k=1^3e^-50p_kdfrac(50p_k)^i_ki_k!$$
where $p_k$ is the probability that someone buys $k$ tickets, for $k=1,2,3$.
probability probability-distributions stochastic-processes poisson-process
asked Mar 28 at 11:25
JB071098JB071098
347212
$endgroup$
add a comment |
$begingroup$
Suppose that the number of women who buy concert tickets follows a Poisson process with rate $30$ women per hour, and similarly the number of men who buy tickets is a Poisson distribution with rate $20$ per hour.
Regardless of gender customers buy
$1$ ticket with probability $1/2$
$2$ tickets with probability $2/5$
$3$ tickets with probabiltiy $1/10$
Letting $N_i$ be the number of customers that buy $i$ tickets in the first hour, find the joint distribution of $(N_1,N_2,N_3)$.
I know that since the number of females $N_F$ and $N_M$ are independent, that their superposition $N_F+N_M$ is a Poisson process with rate $30+20=50$, and also that the arrival times follow an ordered uniform distribution. But I don't what a joint distribution on the vector $(N_1,N_2,N_3)$ would even look like. Any help is appreciated.
Attempt.
Since $N_1+N_2+N_3=N(1)$ is a Poisson process with rate $50$, then
$$N_1sim Poisson(frac12cdot 50)$$
is a Poisson process, and similarly for $N_2$ and $N_3$,
$$N_2sim Poisson(frac25cdot 50)$$
$$N_3sim Poisson(frac110cdot 50)$$
So the joint distribution function is given by
$$P(N_1=i_1,N_2=i_2,N_3=i_3)=prod_k=1^3e^-50p_kdfrac(50p_k)^i_ki_k!$$
where $p_k$ is the probability that someone buys $k$ tickets, for $k=1,2,3$.
probability probability-distributions stochastic-processes poisson-process
asked Mar 28 at 11:25
JB071098JB071098
347212
$endgroup$
1
$begingroup$
This should be called as splitting Poisson process. At the end you should be able to show that the are independent Poisson. See, e.g. math.stackexchange.com/questions/1777427/…
$endgroup$
– BGM
Mar 28 at 12:14
$begingroup$
Could you check to see if I have the right idea? @BGM
$endgroup$
– JB071098
Mar 28 at 12:27
$begingroup$
Yes that should be the desired joint pmf. Now remaining is whether you need to supplement intermediate calculations steps as some sort of proof, or you can just directly quote the result like this.
$endgroup$
– BGM
Mar 29 at 3:36
add a comment |
$begingroup$
Suppose that the number of women who buy concert tickets follows a Poisson process with rate $30$ women per hour, and similarly the number of men who buy tickets is a Poisson distribution with rate $20$ per hour.
Regardless of gender customers buy
$1$ ticket with probability $1/2$
$2$ tickets with probability $2/5$
$3$ tickets with probabiltiy $1/10$
Letting $N_i$ be the number of customers that buy $i$ tickets in the first hour, find the joint distribution of $(N_1,N_2,N_3)$.
I know that since the number of females $N_F$ and $N_M$ are independent, that their superposition $N_F+N_M$ is a Poisson process with rate $30+20=50$, and also that the arrival times follow an ordered uniform distribution. But I don't what a joint distribution on the vector $(N_1,N_2,N_3)$ would even look like. Any help is appreciated.
Attempt.
Since $N_1+N_2+N_3=N(1)$ is a Poisson process with rate $50$, then
$$N_1sim Poisson(frac12cdot 50)$$
is a Poisson process, and similarly for $N_2$ and $N_3$,
$$N_2sim Poisson(frac25cdot 50)$$
$$N_3sim Poisson(frac110cdot 50)$$
So the joint distribution function is given by
$$P(N_1=i_1,N_2=i_2,N_3=i_3)=prod_k=1^3e^-50p_kdfrac(50p_k)^i_ki_k!$$
where $p_k$ is the probability that someone buys $k$ tickets, for $k=1,2,3$.
probability probability-distributions stochastic-processes poisson-process
asked Mar 28 at 11:25
JB071098JB071098
347212
$endgroup$
Suppose that the number of women who buy concert tickets follows a Poisson process with rate $30$ women per hour, and similarly the number of men who buy tickets is a Poisson distribution with rate $20$ per hour.
Regardless of gender customers buy
$1$ ticket with probability $1/2$
$2$ tickets with probability $2/5$
$3$ tickets with probabiltiy $1/10$
Letting $N_i$ be the number of customers that buy $i$ tickets in the first hour, find the joint distribution of $(N_1,N_2,N_3)$.
I know that since the number of females $N_F$ and $N_M$ are independent, that their superposition $N_F+N_M$ is a Poisson process with rate $30+20=50$, and also that the arrival times follow an ordered uniform distribution. But I don't what a joint distribution on the vector $(N_1,N_2,N_3)$ would even look like. Any help is appreciated.
Attempt.
Since $N_1+N_2+N_3=N(1)$ is a Poisson process with rate $50$, then
$$N_1sim Poisson(frac12cdot 50)$$
is a Poisson process, and similarly for $N_2$ and $N_3$,
$$N_2sim Poisson(frac25cdot 50)$$
$$N_3sim Poisson(frac110cdot 50)$$
So the joint distribution function is given by
$$P(N_1=i_1,N_2=i_2,N_3=i_3)=prod_k=1^3e^-50p_kdfrac(50p_k)^i_ki_k!$$
where $p_k$ is the probability that someone buys $k$ tickets, for $k=1,2,3$.
probability probability-distributions stochastic-processes poisson-process
probability probability-distributions stochastic-processes poisson-process
asked Mar 28 at 11:25
JB071098JB071098
347212
asked Mar 28 at 11:25
JB071098JB071098
347212
edited Mar 28 at 12:27
JB071098
asked Mar 28 at 11:25
JB071098JB071098
347212
asked Mar 28 at 11:25
JB071098JB071098
347212
asked Mar 28 at 11:25
JB071098JB071098
347212
347212
1
$begingroup$
This should be called as splitting Poisson process. At the end you should be able to show that the are independent Poisson. See, e.g. math.stackexchange.com/questions/1777427/…
$endgroup$
– BGM
Mar 28 at 12:14
$begingroup$
Could you check to see if I have the right idea? @BGM
$endgroup$
– JB071098
Mar 28 at 12:27
$begingroup$
Yes that should be the desired joint pmf. Now remaining is whether you need to supplement intermediate calculations steps as some sort of proof, or you can just directly quote the result like this.
$endgroup$
– BGM
Mar 29 at 3:36
add a comment |
1
$begingroup$
This should be called as splitting Poisson process. At the end you should be able to show that the are independent Poisson. See, e.g. math.stackexchange.com/questions/1777427/…
$endgroup$
– BGM
Mar 28 at 12:14
$begingroup$
Could you check to see if I have the right idea? @BGM
$endgroup$
– JB071098
Mar 28 at 12:27
$begingroup$
Yes that should be the desired joint pmf. Now remaining is whether you need to supplement intermediate calculations steps as some sort of proof, or you can just directly quote the result like this.
$endgroup$
– BGM
Mar 29 at 3:36
1
1
$begingroup$
This should be called as splitting Poisson process. At the end you should be able to show that the are independent Poisson. See, e.g. math.stackexchange.com/questions/1777427/…
$endgroup$
– BGM
Mar 28 at 12:14
$begingroup$
This should be called as splitting Poisson process. At the end you should be able to show that the are independent Poisson. See, e.g. math.stackexchange.com/questions/1777427/…
$endgroup$
– BGM
Mar 28 at 12:14
$begingroup$
Could you check to see if I have the right idea? @BGM
$endgroup$
– JB071098
Mar 28 at 12:27
$begingroup$
Could you check to see if I have the right idea? @BGM
$endgroup$
– JB071098
Mar 28 at 12:27
$begingroup$
Yes that should be the desired joint pmf. Now remaining is whether you need to supplement intermediate calculations steps as some sort of proof, or you can just directly quote the result like this.
$endgroup$
– BGM
Mar 29 at 3:36
$begingroup$
Yes that should be the desired joint pmf. Now remaining is whether you need to supplement intermediate calculations steps as some sort of proof, or you can just directly quote the result like this.
$endgroup$
– BGM
Mar 29 at 3:36
add a comment |
0
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$begingroup$
This should be called as splitting Poisson process. At the end you should be able to show that the are independent Poisson. See, e.g. math.stackexchange.com/questions/1777427/…
$endgroup$
– BGM
Mar 28 at 12:14
$begingroup$
Could you check to see if I have the right idea? @BGM
$endgroup$
– JB071098
Mar 28 at 12:27
$begingroup$
Yes that should be the desired joint pmf. Now remaining is whether you need to supplement intermediate calculations steps as some sort of proof, or you can just directly quote the result like this.
$endgroup$
– BGM
Mar 29 at 3:36