Mixing
find the condition distribution of $Y_1 mid Y_1 + Y_2$ when $Y_1,Y_1$ are Poisson distributed
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Suppose that we have to independent Poisson RV:s. $Y_1 sim Po(lambda_1), Y_2 sim Po(lambda_1c) $ for some constant $c$
I want to find the distribution $Y_1mid Y_1 +Y_2 $ : here is my attempt but I don't see how this leads to some known distribution.
First:
$P(Y_1 =y_1,Y_1+Y_2 = t) = P(Y_1 = y_1, Y_2 = t-y_1) = P(Y_1 = y_1) P(Y_2 = t-y_1) $
$P(Y_1=y_1|Y_1 + Y_2 = t) = dfracP(Y_1 = y_1) P(Y_2 = t-y_1)sum_s leq t P(Y_1 = y_1) P(Y_2 = t-s) = dfrace^-lambda_1dfraclambda_1^y_1y_1!e^-lambda_1cdfrac(lambda_1c)^t-y_1(t-y_1)!sum_sleq te^-lambda_1dfraclambda_1^ss!e^-lambda_1cdfrac(lambda_1c)^t-st-s! $
probability poisson-distribution
edited Jul 15 at 22:10
Bernard
110k635103
asked Jul 15 at 21:27
Danny
659719
add a comment |Â
up vote
0
down vote
favorite
Suppose that we have to independent Poisson RV:s. $Y_1 sim Po(lambda_1), Y_2 sim Po(lambda_1c) $ for some constant $c$
I want to find the distribution $Y_1mid Y_1 +Y_2 $ : here is my attempt but I don't see how this leads to some known distribution.
First:
$P(Y_1 =y_1,Y_1+Y_2 = t) = P(Y_1 = y_1, Y_2 = t-y_1) = P(Y_1 = y_1) P(Y_2 = t-y_1) $
$P(Y_1=y_1|Y_1 + Y_2 = t) = dfracP(Y_1 = y_1) P(Y_2 = t-y_1)sum_s leq t P(Y_1 = y_1) P(Y_2 = t-s) = dfrace^-lambda_1dfraclambda_1^y_1y_1!e^-lambda_1cdfrac(lambda_1c)^t-y_1(t-y_1)!sum_sleq te^-lambda_1dfraclambda_1^ss!e^-lambda_1cdfrac(lambda_1c)^t-st-s! $
probability poisson-distribution
edited Jul 15 at 22:10
Bernard
110k635103
asked Jul 15 at 21:27
Danny
659719
Can't you get rid of exponentials? Put $lambda_1$s together? Also avoid using $y_1$ both as a bound and free variable in the same equation.
– Arnaud Mortier
Jul 15 at 21:32
Note: There should be no $y_2$ in the expression. They should be $t-y_1$.
– Graham Kemp
Jul 15 at 21:34
ok i fixed it ..
– Danny
Jul 15 at 21:37
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose that we have to independent Poisson RV:s. $Y_1 sim Po(lambda_1), Y_2 sim Po(lambda_1c) $ for some constant $c$
I want to find the distribution $Y_1mid Y_1 +Y_2 $ : here is my attempt but I don't see how this leads to some known distribution.
First:
$P(Y_1 =y_1,Y_1+Y_2 = t) = P(Y_1 = y_1, Y_2 = t-y_1) = P(Y_1 = y_1) P(Y_2 = t-y_1) $
$P(Y_1=y_1|Y_1 + Y_2 = t) = dfracP(Y_1 = y_1) P(Y_2 = t-y_1)sum_s leq t P(Y_1 = y_1) P(Y_2 = t-s) = dfrace^-lambda_1dfraclambda_1^y_1y_1!e^-lambda_1cdfrac(lambda_1c)^t-y_1(t-y_1)!sum_sleq te^-lambda_1dfraclambda_1^ss!e^-lambda_1cdfrac(lambda_1c)^t-st-s! $
probability poisson-distribution
edited Jul 15 at 22:10
Bernard
110k635103
asked Jul 15 at 21:27
Danny
659719
Suppose that we have to independent Poisson RV:s. $Y_1 sim Po(lambda_1), Y_2 sim Po(lambda_1c) $ for some constant $c$
I want to find the distribution $Y_1mid Y_1 +Y_2 $ : here is my attempt but I don't see how this leads to some known distribution.
First:
$P(Y_1 =y_1,Y_1+Y_2 = t) = P(Y_1 = y_1, Y_2 = t-y_1) = P(Y_1 = y_1) P(Y_2 = t-y_1) $
$P(Y_1=y_1|Y_1 + Y_2 = t) = dfracP(Y_1 = y_1) P(Y_2 = t-y_1)sum_s leq t P(Y_1 = y_1) P(Y_2 = t-s) = dfrace^-lambda_1dfraclambda_1^y_1y_1!e^-lambda_1cdfrac(lambda_1c)^t-y_1(t-y_1)!sum_sleq te^-lambda_1dfraclambda_1^ss!e^-lambda_1cdfrac(lambda_1c)^t-st-s! $
probability poisson-distribution
edited Jul 15 at 22:10
Bernard
110k635103
asked Jul 15 at 21:27
Danny
659719
edited Jul 15 at 22:10
Bernard
110k635103
edited Jul 15 at 22:10
Bernard
110k635103
edited Jul 15 at 22:10
Bernard
110k635103
110k635103
asked Jul 15 at 21:27
Danny
659719
asked Jul 15 at 21:27
Danny
659719
asked Jul 15 at 21:27
Danny
659719
659719
Can't you get rid of exponentials? Put $lambda_1$s together? Also avoid using $y_1$ both as a bound and free variable in the same equation.
– Arnaud Mortier
Jul 15 at 21:32
Note: There should be no $y_2$ in the expression. They should be $t-y_1$.
– Graham Kemp
Jul 15 at 21:34
ok i fixed it ..
– Danny
Jul 15 at 21:37
add a comment |Â
Can't you get rid of exponentials? Put $lambda_1$s together? Also avoid using $y_1$ both as a bound and free variable in the same equation.
– Arnaud Mortier
Jul 15 at 21:32
Note: There should be no $y_2$ in the expression. They should be $t-y_1$.
– Graham Kemp
Jul 15 at 21:34
ok i fixed it ..
– Danny
Jul 15 at 21:37
Can't you get rid of exponentials? Put $lambda_1$s together? Also avoid using $y_1$ both as a bound and free variable in the same equation.
– Arnaud Mortier
Jul 15 at 21:32
Can't you get rid of exponentials? Put $lambda_1$s together? Also avoid using $y_1$ both as a bound and free variable in the same equation.
– Arnaud Mortier
Jul 15 at 21:32
Note: There should be no $y_2$ in the expression. They should be $t-y_1$.
– Graham Kemp
Jul 15 at 21:34
Note: There should be no $y_2$ in the expression. They should be $t-y_1$.
– Graham Kemp
Jul 15 at 21:34
ok i fixed it ..
– Danny
Jul 15 at 21:37
ok i fixed it ..
– Danny
Jul 15 at 21:37
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
$Y_1+Y_2simmathcal Ptextois((1+c)lambda_1)$ because you have arrivals occuring independently at constant rates $lambda_1$ and $clambda_1$ in two independent Poison processes; so arrivals are occuring independently at constant rate $(1+c)lambda_1$ in a Poison process.
Therefore:
$$sum_s=0^t dfrace^lambda_1lambda_1^ss!dfrace^clambda_1(clambda_1)^t-s(t-s)!~=~ dfrace^(1+c)lambda_1((1+c)lambda_1)^tt!$$
answered Jul 15 at 21:47
Graham Kemp
80.1k43275
add a comment |Â
up vote
1
down vote
The situation can be viewed as one process with parameter $(1+c)lambda_1$ in which the elements are independently randomly marked as members of $Y_1$ with probability $frac 11+c$ and as members of $Y_2$ with probability $frac c1+c$. Knowing how many elements the overall process produced doesn't change this fact. Thus, if you know that $Y_1+Y_2=t$, then $Y_1$ is distributed according to $mathsfBinomialleft(t,frac11+cright)$.
answered Jul 15 at 22:05
joriki
164k10180328
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$Y_1+Y_2simmathcal Ptextois((1+c)lambda_1)$ because you have arrivals occuring independently at constant rates $lambda_1$ and $clambda_1$ in two independent Poison processes; so arrivals are occuring independently at constant rate $(1+c)lambda_1$ in a Poison process.
Therefore:
$$sum_s=0^t dfrace^lambda_1lambda_1^ss!dfrace^clambda_1(clambda_1)^t-s(t-s)!~=~ dfrace^(1+c)lambda_1((1+c)lambda_1)^tt!$$
answered Jul 15 at 21:47
Graham Kemp
80.1k43275
add a comment |Â
up vote
1
down vote
accepted
$Y_1+Y_2simmathcal Ptextois((1+c)lambda_1)$ because you have arrivals occuring independently at constant rates $lambda_1$ and $clambda_1$ in two independent Poison processes; so arrivals are occuring independently at constant rate $(1+c)lambda_1$ in a Poison process.
Therefore:
$$sum_s=0^t dfrace^lambda_1lambda_1^ss!dfrace^clambda_1(clambda_1)^t-s(t-s)!~=~ dfrace^(1+c)lambda_1((1+c)lambda_1)^tt!$$
answered Jul 15 at 21:47
Graham Kemp
80.1k43275
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$Y_1+Y_2simmathcal Ptextois((1+c)lambda_1)$ because you have arrivals occuring independently at constant rates $lambda_1$ and $clambda_1$ in two independent Poison processes; so arrivals are occuring independently at constant rate $(1+c)lambda_1$ in a Poison process.
Therefore:
$$sum_s=0^t dfrace^lambda_1lambda_1^ss!dfrace^clambda_1(clambda_1)^t-s(t-s)!~=~ dfrace^(1+c)lambda_1((1+c)lambda_1)^tt!$$
answered Jul 15 at 21:47
Graham Kemp
80.1k43275
$Y_1+Y_2simmathcal Ptextois((1+c)lambda_1)$ because you have arrivals occuring independently at constant rates $lambda_1$ and $clambda_1$ in two independent Poison processes; so arrivals are occuring independently at constant rate $(1+c)lambda_1$ in a Poison process.
Therefore:
$$sum_s=0^t dfrace^lambda_1lambda_1^ss!dfrace^clambda_1(clambda_1)^t-s(t-s)!~=~ dfrace^(1+c)lambda_1((1+c)lambda_1)^tt!$$
answered Jul 15 at 21:47
Graham Kemp
80.1k43275
answered Jul 15 at 21:47
Graham Kemp
80.1k43275
answered Jul 15 at 21:47
Graham Kemp
80.1k43275
answered Jul 15 at 21:47
Graham Kemp
80.1k43275
80.1k43275
add a comment |Â
add a comment |Â
up vote
1
down vote
The situation can be viewed as one process with parameter $(1+c)lambda_1$ in which the elements are independently randomly marked as members of $Y_1$ with probability $frac 11+c$ and as members of $Y_2$ with probability $frac c1+c$. Knowing how many elements the overall process produced doesn't change this fact. Thus, if you know that $Y_1+Y_2=t$, then $Y_1$ is distributed according to $mathsfBinomialleft(t,frac11+cright)$.
answered Jul 15 at 22:05
joriki
164k10180328
add a comment |Â
up vote
1
down vote
The situation can be viewed as one process with parameter $(1+c)lambda_1$ in which the elements are independently randomly marked as members of $Y_1$ with probability $frac 11+c$ and as members of $Y_2$ with probability $frac c1+c$. Knowing how many elements the overall process produced doesn't change this fact. Thus, if you know that $Y_1+Y_2=t$, then $Y_1$ is distributed according to $mathsfBinomialleft(t,frac11+cright)$.
answered Jul 15 at 22:05
joriki
164k10180328
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The situation can be viewed as one process with parameter $(1+c)lambda_1$ in which the elements are independently randomly marked as members of $Y_1$ with probability $frac 11+c$ and as members of $Y_2$ with probability $frac c1+c$. Knowing how many elements the overall process produced doesn't change this fact. Thus, if you know that $Y_1+Y_2=t$, then $Y_1$ is distributed according to $mathsfBinomialleft(t,frac11+cright)$.
answered Jul 15 at 22:05
joriki
164k10180328
The situation can be viewed as one process with parameter $(1+c)lambda_1$ in which the elements are independently randomly marked as members of $Y_1$ with probability $frac 11+c$ and as members of $Y_2$ with probability $frac c1+c$. Knowing how many elements the overall process produced doesn't change this fact. Thus, if you know that $Y_1+Y_2=t$, then $Y_1$ is distributed according to $mathsfBinomialleft(t,frac11+cright)$.
answered Jul 15 at 22:05
joriki
164k10180328
answered Jul 15 at 22:05
joriki
164k10180328
answered Jul 15 at 22:05
joriki
164k10180328
answered Jul 15 at 22:05
joriki
164k10180328
164k10180328
add a comment |Â
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%2f2852863%2ffind-the-condition-distribution-of-y-1-mid-y-1-y-2-when-y-1-y-1-are-poiss%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
Can't you get rid of exponentials? Put $lambda_1$s together? Also avoid using $y_1$ both as a bound and free variable in the same equation.
– Arnaud Mortier
Jul 15 at 21:32
Note: There should be no $y_2$ in the expression. They should be $t-y_1$.
– Graham Kemp
Jul 15 at 21:34
ok i fixed it ..
– Danny
Jul 15 at 21:37