Let $X,Y$ independent random variables such that $XsimtextPoisson(lambda)$, and $YsimtextPoisson(mu)$.

Clash Royale CLAN TAG#URR8PPP

up vote
0
down vote

favorite

Let $X,Y$ independent random variables such that $XsimoperatornamePoisson(lambda)$, and
$Ysim operatornamePoisson(mu)$. Let $Z=X+Y$ find the probability function of $X$ and $Z$.

Using the idea of the user @Surb. i have:

$mathbb PX=k,Z=m=mathbb PX=k, X+Y=m=mathbb PX=k, Y=m-k=P(X=k)P(Y=m-k)=e^-lambdafraclambda^kk!e^-mufracmu^m-k(m-k)!=e^-(lambda +mu)fraclambda^k mu^m-kk!(m-k)!$

Then the probability function of $Z$ is

$p(m,k)=begincases
e^-(lambda +mu)fraclambda^k mu^m-kk!(m-k)! && textif& mleq k \
0 && textif & m>k
endcases$

probability

share|cite|improve this question

edited Jul 14 at 14:24

asked Jul 14 at 13:27

Bvss12

1,609516

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You have this only when $mgeq k$. When $m<k$ then $mathbb PX=k, Z=m=0$
  – Surb
  Jul 14 at 14:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks @Surb i put the probability function in my question
  – Bvss12
  Jul 14 at 14:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, it's correct.
  – Surb
  Jul 14 at 14:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for all, very glad for you help @Surb have a great day
  – Bvss12
  Jul 14 at 14:26
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

Let $X,Y$ independent random variables such that $XsimoperatornamePoisson(lambda)$, and
$Ysim operatornamePoisson(mu)$. Let $Z=X+Y$ find the probability function of $X$ and $Z$.

Using the idea of the user @Surb. i have:

$mathbb PX=k,Z=m=mathbb PX=k, X+Y=m=mathbb PX=k, Y=m-k=P(X=k)P(Y=m-k)=e^-lambdafraclambda^kk!e^-mufracmu^m-k(m-k)!=e^-(lambda +mu)fraclambda^k mu^m-kk!(m-k)!$

Then the probability function of $Z$ is

$p(m,k)=begincases
e^-(lambda +mu)fraclambda^k mu^m-kk!(m-k)! && textif& mleq k \
0 && textif & m>k
endcases$

probability

share|cite|improve this question

edited Jul 14 at 14:24

asked Jul 14 at 13:27

Bvss12

1,609516

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You have this only when $mgeq k$. When $m<k$ then $mathbb PX=k, Z=m=0$
  – Surb
  Jul 14 at 14:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks @Surb i put the probability function in my question
  – Bvss12
  Jul 14 at 14:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, it's correct.
  – Surb
  Jul 14 at 14:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for all, very glad for you help @Surb have a great day
  – Bvss12
  Jul 14 at 14:26
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let $X,Y$ independent random variables such that $XsimoperatornamePoisson(lambda)$, and
$Ysim operatornamePoisson(mu)$. Let $Z=X+Y$ find the probability function of $X$ and $Z$.

Using the idea of the user @Surb. i have:

$mathbb PX=k,Z=m=mathbb PX=k, X+Y=m=mathbb PX=k, Y=m-k=P(X=k)P(Y=m-k)=e^-lambdafraclambda^kk!e^-mufracmu^m-k(m-k)!=e^-(lambda +mu)fraclambda^k mu^m-kk!(m-k)!$

Then the probability function of $Z$ is

$p(m,k)=begincases
e^-(lambda +mu)fraclambda^k mu^m-kk!(m-k)! && textif& mleq k \
0 && textif & m>k
endcases$

probability

share|cite|improve this question

edited Jul 14 at 14:24

asked Jul 14 at 13:27

Bvss12

1,609516

Let $X,Y$ independent random variables such that $XsimoperatornamePoisson(lambda)$, and
$Ysim operatornamePoisson(mu)$. Let $Z=X+Y$ find the probability function of $X$ and $Z$.

Using the idea of the user @Surb. i have:

$mathbb PX=k,Z=m=mathbb PX=k, X+Y=m=mathbb PX=k, Y=m-k=P(X=k)P(Y=m-k)=e^-lambdafraclambda^kk!e^-mufracmu^m-k(m-k)!=e^-(lambda +mu)fraclambda^k mu^m-kk!(m-k)!$

Then the probability function of $Z$ is

$p(m,k)=begincases
e^-(lambda +mu)fraclambda^k mu^m-kk!(m-k)! && textif& mleq k \
0 && textif & m>k
endcases$

probability

share|cite|improve this question

edited Jul 14 at 14:24

asked Jul 14 at 13:27

Bvss12

1,609516

share|cite|improve this question

share|cite|improve this question

edited Jul 14 at 14:24

edited Jul 14 at 14:24

edited Jul 14 at 14:24

asked Jul 14 at 13:27

Bvss12

1,609516

asked Jul 14 at 13:27

Bvss12

1,609516

asked Jul 14 at 13:27

Bvss12

1,609516

1,609516

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You have this only when $mgeq k$. When $m<k$ then $mathbb PX=k, Z=m=0$
  – Surb
  Jul 14 at 14:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks @Surb i put the probability function in my question
  – Bvss12
  Jul 14 at 14:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, it's correct.
  – Surb
  Jul 14 at 14:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for all, very glad for you help @Surb have a great day
  – Bvss12
  Jul 14 at 14:26
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You have this only when $mgeq k$. When $m<k$ then $mathbb PX=k, Z=m=0$
  – Surb
  Jul 14 at 14:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks @Surb i put the probability function in my question
  – Bvss12
  Jul 14 at 14:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, it's correct.
  – Surb
  Jul 14 at 14:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for all, very glad for you help @Surb have a great day
  – Bvss12
  Jul 14 at 14:26
  

  
  
  
  

You have this only when $mgeq k$. When $m<k$ then $mathbb PX=k, Z=m=0$
– Surb
Jul 14 at 14:17

You have this only when $mgeq k$. When $m<k$ then $mathbb PX=k, Z=m=0$
– Surb
Jul 14 at 14:17

Thanks @Surb i put the probability function in my question
– Bvss12
Jul 14 at 14:25

Thanks @Surb i put the probability function in my question
– Bvss12
Jul 14 at 14:25

Yes, it's correct.
– Surb
Jul 14 at 14:25

Yes, it's correct.
– Surb
Jul 14 at 14:25

Thanks for all, very glad for you help @Surb have a great day
– Bvss12
Jul 14 at 14:26

Thanks for all, very glad for you help @Surb have a great day
– Bvss12
Jul 14 at 14:26

add a comment | 

2 Answers
2

active

oldest

votes

up vote
1
down vote



accepted

Hint

Let $m,kinmathbb N$ s.t. $mgeq k$. Then

$$mathbb PX=k,Z=m=mathbb PX=k, X+Y=m=mathbb PX=k, Y=m-k=...$$

I let you conclude when $m<k$.

share|cite|improve this answer

edited Jul 14 at 13:35

answered Jul 14 at 13:31

Surb

36.3k84274

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then $P(Z=z)=e^(lambda +mu)frac(lambda +mu)^zz!$
  – Bvss12
  Jul 14 at 13:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I edited my answer. @Bvss12
  – Surb
  Jul 14 at 13:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, then i need know when $m<k$ and with that i can find the probability function. no?
  – Bvss12
  Jul 14 at 13:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes exactly. But don't look to far for the case $m<k$ because it's really obvious.@Bvss12
  – Surb
  Jul 14 at 13:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay, i go to try solve and comment here for any question. Thanks for all!
  – Bvss12
  Jul 14 at 13:39
  

  
  
  
  

 | 
show 1 more comment

up vote
0
down vote

Since the approach using $P(Z=n)=sum_k=0^n P(X=k)P(Y=n-k)$ has already been covered, I'll discuss a different one that requires less combinatorics. Define $$G_X(t):=mathbbEt^X=sum_kge 0e^-lambdafrac(lambda t)^kk!=e^lambda (t-1).$$This power series has $t^k$ coefficient $P(X=k)$, so is called a probability generating function. Clearly, $G_Z=mathbbEt^X t^Y=G_X G_Y=e^(lambda+mu)(t-1)$, i.e. $Z$ is Poisson-distributed with parameter $lambda+mu$.

share|cite|improve this answer

answered Jul 14 at 14:09

J.G.

13.2k11424

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Hint

Let $m,kinmathbb N$ s.t. $mgeq k$. Then

$$mathbb PX=k,Z=m=mathbb PX=k, X+Y=m=mathbb PX=k, Y=m-k=...$$

I let you conclude when $m<k$.

share|cite|improve this answer

edited Jul 14 at 13:35

answered Jul 14 at 13:31

Surb

36.3k84274

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then $P(Z=z)=e^(lambda +mu)frac(lambda +mu)^zz!$
  – Bvss12
  Jul 14 at 13:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I edited my answer. @Bvss12
  – Surb
  Jul 14 at 13:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, then i need know when $m<k$ and with that i can find the probability function. no?
  – Bvss12
  Jul 14 at 13:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes exactly. But don't look to far for the case $m<k$ because it's really obvious.@Bvss12
  – Surb
  Jul 14 at 13:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay, i go to try solve and comment here for any question. Thanks for all!
  – Bvss12
  Jul 14 at 13:39
  

  
  
  
  

 | 
show 1 more comment

up vote
1
down vote



accepted

Hint

Let $m,kinmathbb N$ s.t. $mgeq k$. Then

$$mathbb PX=k,Z=m=mathbb PX=k, X+Y=m=mathbb PX=k, Y=m-k=...$$

I let you conclude when $m<k$.

share|cite|improve this answer

edited Jul 14 at 13:35

answered Jul 14 at 13:31

Surb

36.3k84274

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then $P(Z=z)=e^(lambda +mu)frac(lambda +mu)^zz!$
  – Bvss12
  Jul 14 at 13:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I edited my answer. @Bvss12
  – Surb
  Jul 14 at 13:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, then i need know when $m<k$ and with that i can find the probability function. no?
  – Bvss12
  Jul 14 at 13:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes exactly. But don't look to far for the case $m<k$ because it's really obvious.@Bvss12
  – Surb
  Jul 14 at 13:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay, i go to try solve and comment here for any question. Thanks for all!
  – Bvss12
  Jul 14 at 13:39
  

  
  
  
  

 | 
show 1 more comment

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Hint

Let $m,kinmathbb N$ s.t. $mgeq k$. Then

$$mathbb PX=k,Z=m=mathbb PX=k, X+Y=m=mathbb PX=k, Y=m-k=...$$

I let you conclude when $m<k$.

share|cite|improve this answer

edited Jul 14 at 13:35

answered Jul 14 at 13:31

Surb

36.3k84274

Hint

Let $m,kinmathbb N$ s.t. $mgeq k$. Then

$$mathbb PX=k,Z=m=mathbb PX=k, X+Y=m=mathbb PX=k, Y=m-k=...$$

I let you conclude when $m<k$.

share|cite|improve this answer

edited Jul 14 at 13:35

answered Jul 14 at 13:31

Surb

36.3k84274

share|cite|improve this answer

share|cite|improve this answer

edited Jul 14 at 13:35

edited Jul 14 at 13:35

edited Jul 14 at 13:35

answered Jul 14 at 13:31

Surb

36.3k84274

answered Jul 14 at 13:31

Surb

36.3k84274

answered Jul 14 at 13:31

Surb

36.3k84274

36.3k84274

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then $P(Z=z)=e^(lambda +mu)frac(lambda +mu)^zz!$
  – Bvss12
  Jul 14 at 13:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I edited my answer. @Bvss12
  – Surb
  Jul 14 at 13:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, then i need know when $m<k$ and with that i can find the probability function. no?
  – Bvss12
  Jul 14 at 13:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes exactly. But don't look to far for the case $m<k$ because it's really obvious.@Bvss12
  – Surb
  Jul 14 at 13:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay, i go to try solve and comment here for any question. Thanks for all!
  – Bvss12
  Jul 14 at 13:39
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then $P(Z=z)=e^(lambda +mu)frac(lambda +mu)^zz!$
  – Bvss12
  Jul 14 at 13:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I edited my answer. @Bvss12
  – Surb
  Jul 14 at 13:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, then i need know when $m<k$ and with that i can find the probability function. no?
  – Bvss12
  Jul 14 at 13:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes exactly. But don't look to far for the case $m<k$ because it's really obvious.@Bvss12
  – Surb
  Jul 14 at 13:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay, i go to try solve and comment here for any question. Thanks for all!
  – Bvss12
  Jul 14 at 13:39
  

  
  
  
  

Then $P(Z=z)=e^(lambda +mu)frac(lambda +mu)^zz!$
– Bvss12
Jul 14 at 13:34

Then $P(Z=z)=e^(lambda +mu)frac(lambda +mu)^zz!$
– Bvss12
Jul 14 at 13:34

I edited my answer. @Bvss12
– Surb
Jul 14 at 13:36

I edited my answer. @Bvss12
– Surb
Jul 14 at 13:36

Thanks, then i need know when $m<k$ and with that i can find the probability function. no?
– Bvss12
Jul 14 at 13:37

Thanks, then i need know when $m<k$ and with that i can find the probability function. no?
– Bvss12
Jul 14 at 13:37

Yes exactly. But don't look to far for the case $m<k$ because it's really obvious.@Bvss12
– Surb
Jul 14 at 13:38

Yes exactly. But don't look to far for the case $m<k$ because it's really obvious.@Bvss12
– Surb
Jul 14 at 13:38

Okay, i go to try solve and comment here for any question. Thanks for all!
– Bvss12
Jul 14 at 13:39

Okay, i go to try solve and comment here for any question. Thanks for all!
– Bvss12
Jul 14 at 13:39

 | 
show 1 more comment

up vote
0
down vote

Since the approach using $P(Z=n)=sum_k=0^n P(X=k)P(Y=n-k)$ has already been covered, I'll discuss a different one that requires less combinatorics. Define $$G_X(t):=mathbbEt^X=sum_kge 0e^-lambdafrac(lambda t)^kk!=e^lambda (t-1).$$This power series has $t^k$ coefficient $P(X=k)$, so is called a probability generating function. Clearly, $G_Z=mathbbEt^X t^Y=G_X G_Y=e^(lambda+mu)(t-1)$, i.e. $Z$ is Poisson-distributed with parameter $lambda+mu$.

share|cite|improve this answer

answered Jul 14 at 14:09

J.G.

13.2k11424

add a comment | 

up vote
0
down vote

Since the approach using $P(Z=n)=sum_k=0^n P(X=k)P(Y=n-k)$ has already been covered, I'll discuss a different one that requires less combinatorics. Define $$G_X(t):=mathbbEt^X=sum_kge 0e^-lambdafrac(lambda t)^kk!=e^lambda (t-1).$$This power series has $t^k$ coefficient $P(X=k)$, so is called a probability generating function. Clearly, $G_Z=mathbbEt^X t^Y=G_X G_Y=e^(lambda+mu)(t-1)$, i.e. $Z$ is Poisson-distributed with parameter $lambda+mu$.

share|cite|improve this answer

answered Jul 14 at 14:09

J.G.

13.2k11424

add a comment | 

up vote
0
down vote

up vote
0
down vote

Since the approach using $P(Z=n)=sum_k=0^n P(X=k)P(Y=n-k)$ has already been covered, I'll discuss a different one that requires less combinatorics. Define $$G_X(t):=mathbbEt^X=sum_kge 0e^-lambdafrac(lambda t)^kk!=e^lambda (t-1).$$This power series has $t^k$ coefficient $P(X=k)$, so is called a probability generating function. Clearly, $G_Z=mathbbEt^X t^Y=G_X G_Y=e^(lambda+mu)(t-1)$, i.e. $Z$ is Poisson-distributed with parameter $lambda+mu$.

share|cite|improve this answer

answered Jul 14 at 14:09

J.G.

13.2k11424

Since the approach using $P(Z=n)=sum_k=0^n P(X=k)P(Y=n-k)$ has already been covered, I'll discuss a different one that requires less combinatorics. Define $$G_X(t):=mathbbEt^X=sum_kge 0e^-lambdafrac(lambda t)^kk!=e^lambda (t-1).$$This power series has $t^k$ coefficient $P(X=k)$, so is called a probability generating function. Clearly, $G_Z=mathbbEt^X t^Y=G_X G_Y=e^(lambda+mu)(t-1)$, i.e. $Z$ is Poisson-distributed with parameter $lambda+mu$.

share|cite|improve this answer

answered Jul 14 at 14:09

J.G.

13.2k11424

share|cite|improve this answer

share|cite|improve this answer

answered Jul 14 at 14:09

J.G.

13.2k11424

answered Jul 14 at 14:09

J.G.

13.2k11424

answered Jul 14 at 14:09

J.G.

13.2k11424

13.2k11424

add a comment | 

add a comment | 

 

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