Find the probability $P(X<Y)$.
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Suppose $X$ and $Y$ be two independent Poisson $(lambda)$ random variables. Find $P(X<Y)$.
My attempt $:$
beginalign*P(X<Y) &= sumlimits_x=0^infty sumlimits_y=x+1^infty P(X=x,Y=y) \
&= sumlimits_x=0^infty sumlimits_y=x+1^infty P(X=x) P(Y=y) \
&= e^-2lambda sumlimits_x=0^infty frac lambda^x x! sumlimits_y=x+1^infty frac lambda^y y! \
&= e^-2lambda sumlimits_x=0^inftyleft( frace^-lambdalambda^xx! - fraclambda^xx! sumlimits_y=0^x frac lambda^yy! right) \
&= e^-2lambda - e^-2lambda sumlimits_x=0^infty left( frac lambda^x x! sumlimits_y=0^x frac lambda^y y! right)
endalign*
Now how do I calculate $$sumlimits_x=0^infty left( frac lambda^x x! sumlimits_y=0^x frac lambda^y y! right)$$ Please help me in this regard.
Thank you very much.
probability probability-distributions poisson-distribution
asked Sep 8 at 8:34
Dbchatto67
34313
 |Â
show 3 more comments
up vote
2
down vote
favorite
Suppose $X$ and $Y$ be two independent Poisson $(lambda)$ random variables. Find $P(X<Y)$.
My attempt $:$
beginalign*P(X<Y) &= sumlimits_x=0^infty sumlimits_y=x+1^infty P(X=x,Y=y) \
&= sumlimits_x=0^infty sumlimits_y=x+1^infty P(X=x) P(Y=y) \
&= e^-2lambda sumlimits_x=0^infty frac lambda^x x! sumlimits_y=x+1^infty frac lambda^y y! \
&= e^-2lambda sumlimits_x=0^inftyleft( frace^-lambdalambda^xx! - fraclambda^xx! sumlimits_y=0^x frac lambda^yy! right) \
&= e^-2lambda - e^-2lambda sumlimits_x=0^infty left( frac lambda^x x! sumlimits_y=0^x frac lambda^y y! right)
endalign*
Now how do I calculate $$sumlimits_x=0^infty left( frac lambda^x x! sumlimits_y=0^x frac lambda^y y! right)$$ Please help me in this regard.
Thank you very much.
probability probability-distributions poisson-distribution
asked Sep 8 at 8:34
Dbchatto67
34313
can you edit your question there is some error in your latex formatting
– Deepesh Meena
Sep 8 at 8:42
Please somebody point out where my error was.
– Dbchatto67
Sep 8 at 9:03
please check if the edit is correct.
– Siong Thye Goh
Sep 8 at 9:15
3
Simpler: By symmetry, $P(X<Y)=P(X>Y)$ hence $P(X<Y)=frac12-frac12P(X=Y)$ with $$P(X=Y)=e^-2lambdasum_n=0^inftyfraclambda^2n(n!)^2$$ for which there is no simpler expression (except, rather circularly, by identifying the sum as a value of a special function defined as the sum of this series).
– Did
Sep 8 at 9:33
1
@Mark The page you link to in your answer below is actually a good start to see the problem: note how people there call "Hint" or "Answer" the mere mention of the name of a series which the OP already computed on their own.
– Did
Sep 8 at 10:10
 |Â
show 3 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose $X$ and $Y$ be two independent Poisson $(lambda)$ random variables. Find $P(X<Y)$.
My attempt $:$
beginalign*P(X<Y) &= sumlimits_x=0^infty sumlimits_y=x+1^infty P(X=x,Y=y) \
&= sumlimits_x=0^infty sumlimits_y=x+1^infty P(X=x) P(Y=y) \
&= e^-2lambda sumlimits_x=0^infty frac lambda^x x! sumlimits_y=x+1^infty frac lambda^y y! \
&= e^-2lambda sumlimits_x=0^inftyleft( frace^-lambdalambda^xx! - fraclambda^xx! sumlimits_y=0^x frac lambda^yy! right) \
&= e^-2lambda - e^-2lambda sumlimits_x=0^infty left( frac lambda^x x! sumlimits_y=0^x frac lambda^y y! right)
endalign*
Now how do I calculate $$sumlimits_x=0^infty left( frac lambda^x x! sumlimits_y=0^x frac lambda^y y! right)$$ Please help me in this regard.
Thank you very much.
probability probability-distributions poisson-distribution
asked Sep 8 at 8:34
Dbchatto67
34313
Suppose $X$ and $Y$ be two independent Poisson $(lambda)$ random variables. Find $P(X<Y)$.
My attempt $:$
beginalign*P(X<Y) &= sumlimits_x=0^infty sumlimits_y=x+1^infty P(X=x,Y=y) \
&= sumlimits_x=0^infty sumlimits_y=x+1^infty P(X=x) P(Y=y) \
&= e^-2lambda sumlimits_x=0^infty frac lambda^x x! sumlimits_y=x+1^infty frac lambda^y y! \
&= e^-2lambda sumlimits_x=0^inftyleft( frace^-lambdalambda^xx! - fraclambda^xx! sumlimits_y=0^x frac lambda^yy! right) \
&= e^-2lambda - e^-2lambda sumlimits_x=0^infty left( frac lambda^x x! sumlimits_y=0^x frac lambda^y y! right)
endalign*
Now how do I calculate $$sumlimits_x=0^infty left( frac lambda^x x! sumlimits_y=0^x frac lambda^y y! right)$$ Please help me in this regard.
Thank you very much.
probability probability-distributions poisson-distribution
probability probability-distributions poisson-distribution
asked Sep 8 at 8:34
Dbchatto67
34313
asked Sep 8 at 8:34
Dbchatto67
34313
edited Sep 8 at 9:22
asked Sep 8 at 8:34
Dbchatto67
34313
asked Sep 8 at 8:34
Dbchatto67
34313
asked Sep 8 at 8:34
Dbchatto67
34313
34313
can you edit your question there is some error in your latex formatting
– Deepesh Meena
Sep 8 at 8:42
Please somebody point out where my error was.
– Dbchatto67
Sep 8 at 9:03
please check if the edit is correct.
– Siong Thye Goh
Sep 8 at 9:15
3
Simpler: By symmetry, $P(X<Y)=P(X>Y)$ hence $P(X<Y)=frac12-frac12P(X=Y)$ with $$P(X=Y)=e^-2lambdasum_n=0^inftyfraclambda^2n(n!)^2$$ for which there is no simpler expression (except, rather circularly, by identifying the sum as a value of a special function defined as the sum of this series).
– Did
Sep 8 at 9:33
1
@Mark The page you link to in your answer below is actually a good start to see the problem: note how people there call "Hint" or "Answer" the mere mention of the name of a series which the OP already computed on their own.
– Did
Sep 8 at 10:10
 |Â
show 3 more comments
can you edit your question there is some error in your latex formatting
– Deepesh Meena
Sep 8 at 8:42
Please somebody point out where my error was.
– Dbchatto67
Sep 8 at 9:03
please check if the edit is correct.
– Siong Thye Goh
Sep 8 at 9:15
3
Simpler: By symmetry, $P(X<Y)=P(X>Y)$ hence $P(X<Y)=frac12-frac12P(X=Y)$ with $$P(X=Y)=e^-2lambdasum_n=0^inftyfraclambda^2n(n!)^2$$ for which there is no simpler expression (except, rather circularly, by identifying the sum as a value of a special function defined as the sum of this series).
– Did
Sep 8 at 9:33
1
@Mark The page you link to in your answer below is actually a good start to see the problem: note how people there call "Hint" or "Answer" the mere mention of the name of a series which the OP already computed on their own.
– Did
Sep 8 at 10:10
can you edit your question there is some error in your latex formatting
– Deepesh Meena
Sep 8 at 8:42
can you edit your question there is some error in your latex formatting
– Deepesh Meena
Sep 8 at 8:42
Please somebody point out where my error was.
– Dbchatto67
Sep 8 at 9:03
Please somebody point out where my error was.
– Dbchatto67
Sep 8 at 9:03
please check if the edit is correct.
– Siong Thye Goh
Sep 8 at 9:15
please check if the edit is correct.
– Siong Thye Goh
Sep 8 at 9:15
3
3
Simpler: By symmetry, $P(X<Y)=P(X>Y)$ hence $P(X<Y)=frac12-frac12P(X=Y)$ with $$P(X=Y)=e^-2lambdasum_n=0^inftyfraclambda^2n(n!)^2$$ for which there is no simpler expression (except, rather circularly, by identifying the sum as a value of a special function defined as the sum of this series).
– Did
Sep 8 at 9:33
Simpler: By symmetry, $P(X<Y)=P(X>Y)$ hence $P(X<Y)=frac12-frac12P(X=Y)$ with $$P(X=Y)=e^-2lambdasum_n=0^inftyfraclambda^2n(n!)^2$$ for which there is no simpler expression (except, rather circularly, by identifying the sum as a value of a special function defined as the sum of this series).
– Did
Sep 8 at 9:33
1
1
@Mark The page you link to in your answer below is actually a good start to see the problem: note how people there call "Hint" or "Answer" the mere mention of the name of a series which the OP already computed on their own.
– Did
Sep 8 at 10:10
@Mark The page you link to in your answer below is actually a good start to see the problem: note how people there call "Hint" or "Answer" the mere mention of the name of a series which the OP already computed on their own.
– Did
Sep 8 at 10:10
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
For the special case of $X$ and $Y$ being identically distributed, you have
$$P(X < Y) + P(Y < X) + P(Y = X) = 1$$
$$2 P(X < Y) + P(X = Y) = 1$$
$$ P( X < Y) = 1/2 (1 - P(X = Y))$$
So it reduces to computing $P(X =Y)$ whose computation appears here Probability that two independent Poisson random variables with same paramter are equal
answered Sep 8 at 9:32
Mark
1,84322247
Yeah @Mark I have found it myself just now without seeing your answer.
– Dbchatto67
Sep 8 at 10:10
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
For the special case of $X$ and $Y$ being identically distributed, you have
$$P(X < Y) + P(Y < X) + P(Y = X) = 1$$
$$2 P(X < Y) + P(X = Y) = 1$$
$$ P( X < Y) = 1/2 (1 - P(X = Y))$$
So it reduces to computing $P(X =Y)$ whose computation appears here Probability that two independent Poisson random variables with same paramter are equal
answered Sep 8 at 9:32
Mark
1,84322247
Yeah @Mark I have found it myself just now without seeing your answer.
– Dbchatto67
Sep 8 at 10:10
add a comment |Â
up vote
2
down vote
accepted
For the special case of $X$ and $Y$ being identically distributed, you have
$$P(X < Y) + P(Y < X) + P(Y = X) = 1$$
$$2 P(X < Y) + P(X = Y) = 1$$
$$ P( X < Y) = 1/2 (1 - P(X = Y))$$
So it reduces to computing $P(X =Y)$ whose computation appears here Probability that two independent Poisson random variables with same paramter are equal
answered Sep 8 at 9:32
Mark
1,84322247
Yeah @Mark I have found it myself just now without seeing your answer.
– Dbchatto67
Sep 8 at 10:10
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
For the special case of $X$ and $Y$ being identically distributed, you have
$$P(X < Y) + P(Y < X) + P(Y = X) = 1$$
$$2 P(X < Y) + P(X = Y) = 1$$
$$ P( X < Y) = 1/2 (1 - P(X = Y))$$
So it reduces to computing $P(X =Y)$ whose computation appears here Probability that two independent Poisson random variables with same paramter are equal
answered Sep 8 at 9:32
Mark
1,84322247
For the special case of $X$ and $Y$ being identically distributed, you have
$$P(X < Y) + P(Y < X) + P(Y = X) = 1$$
$$2 P(X < Y) + P(X = Y) = 1$$
$$ P( X < Y) = 1/2 (1 - P(X = Y))$$
So it reduces to computing $P(X =Y)$ whose computation appears here Probability that two independent Poisson random variables with same paramter are equal
answered Sep 8 at 9:32
Mark
1,84322247
answered Sep 8 at 9:32
Mark
1,84322247
answered Sep 8 at 9:32
Mark
1,84322247
answered Sep 8 at 9:32
Mark
1,84322247
1,84322247
Yeah @Mark I have found it myself just now without seeing your answer.
– Dbchatto67
Sep 8 at 10:10
add a comment |Â
Yeah @Mark I have found it myself just now without seeing your answer.
– Dbchatto67
Sep 8 at 10:10
Yeah @Mark I have found it myself just now without seeing your answer.
– Dbchatto67
Sep 8 at 10:10
Yeah @Mark I have found it myself just now without seeing your answer.
– Dbchatto67
Sep 8 at 10:10
add a comment |Â
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can you edit your question there is some error in your latex formatting
– Deepesh Meena
Sep 8 at 8:42
Please somebody point out where my error was.
– Dbchatto67
Sep 8 at 9:03
please check if the edit is correct.
– Siong Thye Goh
Sep 8 at 9:15
3
Simpler: By symmetry, $P(X<Y)=P(X>Y)$ hence $P(X<Y)=frac12-frac12P(X=Y)$ with $$P(X=Y)=e^-2lambdasum_n=0^inftyfraclambda^2n(n!)^2$$ for which there is no simpler expression (except, rather circularly, by identifying the sum as a value of a special function defined as the sum of this series).
– Did
Sep 8 at 9:33
1
@Mark The page you link to in your answer below is actually a good start to see the problem: note how people there call "Hint" or "Answer" the mere mention of the name of a series which the OP already computed on their own.
– Did
Sep 8 at 10:10