$P(2X+2Y=28)$ Poisson Variable Distribution












0












$begingroup$



Let $X$ be a random poisson variable with the parameter $5$, let $Y$ be a random poisson variable with the parameter $10$. $X, Y$ are independent of each other.



What is the probability that $2(X+Y) = 28?$




My Approach
Let $Z=2(X+Y)$



Since $X$ and $Y$ are indepenent variables and and indepedent on each other, $Z$ is a poisson random variable with the parameter $lambda_Z = lambda_X + lambda_Y = 15$.



Therefore we can use the poisson probability function -



$$P(Z=28)=P(2X+2Y=28) = e^{-15} cdot frac{15^{28}}{28!} = 0.00085$$



However I'm afraid I'm wrong here. What do I do with the $2$ here? If I had only $X+Y$ I would be pretty sure I'm right, but I am not sure.



Should I multiplie the result on $2$?



Thank you!






probability random-variables poisson-distribution







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    $$2x+2y=28 iff x+y=14$$
    $endgroup$
    – rlartiga
    Jan 2 at 18:25










  • $begingroup$
    So my final result should be multiplied by 2?
    $endgroup$
    – Alan
    Jan 2 at 18:25






  • 1




    $begingroup$
    no just calculate the probability of $X+Y$ equal to $14$
    $endgroup$
    – rlartiga
    Jan 2 at 18:29










  • $begingroup$
    Or look it in this way $Z=X+Y$ so you are asked to find the probability of $2Z=28$
    $endgroup$
    – rlartiga
    Jan 2 at 18:37










  • $begingroup$
    If $X, Y$ are independent Poisson, then $2(X+Y)$ is not Poisson, since its value is always an even integer (but Poisson variables might be odd).
    $endgroup$
    – Michael
    Jan 2 at 18:42
















0












$begingroup$



Let $X$ be a random poisson variable with the parameter $5$, let $Y$ be a random poisson variable with the parameter $10$. $X, Y$ are independent of each other.



What is the probability that $2(X+Y) = 28?$




My Approach
Let $Z=2(X+Y)$



Since $X$ and $Y$ are indepenent variables and and indepedent on each other, $Z$ is a poisson random variable with the parameter $lambda_Z = lambda_X + lambda_Y = 15$.



Therefore we can use the poisson probability function -



$$P(Z=28)=P(2X+2Y=28) = e^{-15} cdot frac{15^{28}}{28!} = 0.00085$$



However I'm afraid I'm wrong here. What do I do with the $2$ here? If I had only $X+Y$ I would be pretty sure I'm right, but I am not sure.



Should I multiplie the result on $2$?



Thank you!






probability random-variables poisson-distribution







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    $$2x+2y=28 iff x+y=14$$
    $endgroup$
    – rlartiga
    Jan 2 at 18:25










  • $begingroup$
    So my final result should be multiplied by 2?
    $endgroup$
    – Alan
    Jan 2 at 18:25






  • 1




    $begingroup$
    no just calculate the probability of $X+Y$ equal to $14$
    $endgroup$
    – rlartiga
    Jan 2 at 18:29










  • $begingroup$
    Or look it in this way $Z=X+Y$ so you are asked to find the probability of $2Z=28$
    $endgroup$
    – rlartiga
    Jan 2 at 18:37










  • $begingroup$
    If $X, Y$ are independent Poisson, then $2(X+Y)$ is not Poisson, since its value is always an even integer (but Poisson variables might be odd).
    $endgroup$
    – Michael
    Jan 2 at 18:42














0












0








0





$begingroup$



Let $X$ be a random poisson variable with the parameter $5$, let $Y$ be a random poisson variable with the parameter $10$. $X, Y$ are independent of each other.



What is the probability that $2(X+Y) = 28?$




My Approach
Let $Z=2(X+Y)$



Since $X$ and $Y$ are indepenent variables and and indepedent on each other, $Z$ is a poisson random variable with the parameter $lambda_Z = lambda_X + lambda_Y = 15$.



Therefore we can use the poisson probability function -



$$P(Z=28)=P(2X+2Y=28) = e^{-15} cdot frac{15^{28}}{28!} = 0.00085$$



However I'm afraid I'm wrong here. What do I do with the $2$ here? If I had only $X+Y$ I would be pretty sure I'm right, but I am not sure.



Should I multiplie the result on $2$?



Thank you!






probability random-variables poisson-distribution







share|cite|improve this question











$endgroup$





Let $X$ be a random poisson variable with the parameter $5$, let $Y$ be a random poisson variable with the parameter $10$. $X, Y$ are independent of each other.



What is the probability that $2(X+Y) = 28?$




My Approach
Let $Z=2(X+Y)$



Since $X$ and $Y$ are indepenent variables and and indepedent on each other, $Z$ is a poisson random variable with the parameter $lambda_Z = lambda_X + lambda_Y = 15$.



Therefore we can use the poisson probability function -



$$P(Z=28)=P(2X+2Y=28) = e^{-15} cdot frac{15^{28}}{28!} = 0.00085$$



However I'm afraid I'm wrong here. What do I do with the $2$ here? If I had only $X+Y$ I would be pretty sure I'm right, but I am not sure.



Should I multiplie the result on $2$?



Thank you!






probability random-variables poisson-distribution




probability random-variables poisson-distribution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 18:44









rlartiga

3,9351924




3,9351924










asked Jan 2 at 18:15









AlanAlan

1,3841021




1,3841021








  • 2




    $begingroup$
    $$2x+2y=28 iff x+y=14$$
    $endgroup$
    – rlartiga
    Jan 2 at 18:25










  • $begingroup$
    So my final result should be multiplied by 2?
    $endgroup$
    – Alan
    Jan 2 at 18:25






  • 1




    $begingroup$
    no just calculate the probability of $X+Y$ equal to $14$
    $endgroup$
    – rlartiga
    Jan 2 at 18:29










  • $begingroup$
    Or look it in this way $Z=X+Y$ so you are asked to find the probability of $2Z=28$
    $endgroup$
    – rlartiga
    Jan 2 at 18:37










  • $begingroup$
    If $X, Y$ are independent Poisson, then $2(X+Y)$ is not Poisson, since its value is always an even integer (but Poisson variables might be odd).
    $endgroup$
    – Michael
    Jan 2 at 18:42














  • 2




    $begingroup$
    $$2x+2y=28 iff x+y=14$$
    $endgroup$
    – rlartiga
    Jan 2 at 18:25










  • $begingroup$
    So my final result should be multiplied by 2?
    $endgroup$
    – Alan
    Jan 2 at 18:25






  • 1




    $begingroup$
    no just calculate the probability of $X+Y$ equal to $14$
    $endgroup$
    – rlartiga
    Jan 2 at 18:29










  • $begingroup$
    Or look it in this way $Z=X+Y$ so you are asked to find the probability of $2Z=28$
    $endgroup$
    – rlartiga
    Jan 2 at 18:37










  • $begingroup$
    If $X, Y$ are independent Poisson, then $2(X+Y)$ is not Poisson, since its value is always an even integer (but Poisson variables might be odd).
    $endgroup$
    – Michael
    Jan 2 at 18:42








2




2




$begingroup$
$$2x+2y=28 iff x+y=14$$
$endgroup$
– rlartiga
Jan 2 at 18:25




$begingroup$
$$2x+2y=28 iff x+y=14$$
$endgroup$
– rlartiga
Jan 2 at 18:25












$begingroup$
So my final result should be multiplied by 2?
$endgroup$
– Alan
Jan 2 at 18:25




$begingroup$
So my final result should be multiplied by 2?
$endgroup$
– Alan
Jan 2 at 18:25




1




1




$begingroup$
no just calculate the probability of $X+Y$ equal to $14$
$endgroup$
– rlartiga
Jan 2 at 18:29




$begingroup$
no just calculate the probability of $X+Y$ equal to $14$
$endgroup$
– rlartiga
Jan 2 at 18:29












$begingroup$
Or look it in this way $Z=X+Y$ so you are asked to find the probability of $2Z=28$
$endgroup$
– rlartiga
Jan 2 at 18:37




$begingroup$
Or look it in this way $Z=X+Y$ so you are asked to find the probability of $2Z=28$
$endgroup$
– rlartiga
Jan 2 at 18:37












$begingroup$
If $X, Y$ are independent Poisson, then $2(X+Y)$ is not Poisson, since its value is always an even integer (but Poisson variables might be odd).
$endgroup$
– Michael
Jan 2 at 18:42




$begingroup$
If $X, Y$ are independent Poisson, then $2(X+Y)$ is not Poisson, since its value is always an even integer (but Poisson variables might be odd).
$endgroup$
– Michael
Jan 2 at 18:42










1 Answer
1






active

oldest

votes


















1












$begingroup$

If $X$ and $Y$ have Poisson-distribution with rates $lambda_X$, $lambda_Y$ and are independent then $X+Y$ has Poisson-distribution with rate $lambda_X+lambda_Y$.



This can be exploited to find: $$P(2(X+Y)=28)=P(X+Y=14)$$



Observe for $Z:=2(X+Y)$ we have $P(Z=n)=0$ if $n$ is odd, showing that it has not Poisson-distribution.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! got it.
    $endgroup$
    – Alan
    Jan 2 at 18:45










  • $begingroup$
    You are welcome.
    $endgroup$
    – drhab
    Jan 2 at 18:46











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

If $X$ and $Y$ have Poisson-distribution with rates $lambda_X$, $lambda_Y$ and are independent then $X+Y$ has Poisson-distribution with rate $lambda_X+lambda_Y$.



This can be exploited to find: $$P(2(X+Y)=28)=P(X+Y=14)$$



Observe for $Z:=2(X+Y)$ we have $P(Z=n)=0$ if $n$ is odd, showing that it has not Poisson-distribution.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! got it.
    $endgroup$
    – Alan
    Jan 2 at 18:45










  • $begingroup$
    You are welcome.
    $endgroup$
    – drhab
    Jan 2 at 18:46
















1












$begingroup$

If $X$ and $Y$ have Poisson-distribution with rates $lambda_X$, $lambda_Y$ and are independent then $X+Y$ has Poisson-distribution with rate $lambda_X+lambda_Y$.



This can be exploited to find: $$P(2(X+Y)=28)=P(X+Y=14)$$



Observe for $Z:=2(X+Y)$ we have $P(Z=n)=0$ if $n$ is odd, showing that it has not Poisson-distribution.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! got it.
    $endgroup$
    – Alan
    Jan 2 at 18:45










  • $begingroup$
    You are welcome.
    $endgroup$
    – drhab
    Jan 2 at 18:46














1












1








1





$begingroup$

If $X$ and $Y$ have Poisson-distribution with rates $lambda_X$, $lambda_Y$ and are independent then $X+Y$ has Poisson-distribution with rate $lambda_X+lambda_Y$.



This can be exploited to find: $$P(2(X+Y)=28)=P(X+Y=14)$$



Observe for $Z:=2(X+Y)$ we have $P(Z=n)=0$ if $n$ is odd, showing that it has not Poisson-distribution.






share|cite|improve this answer









$endgroup$



If $X$ and $Y$ have Poisson-distribution with rates $lambda_X$, $lambda_Y$ and are independent then $X+Y$ has Poisson-distribution with rate $lambda_X+lambda_Y$.



This can be exploited to find: $$P(2(X+Y)=28)=P(X+Y=14)$$



Observe for $Z:=2(X+Y)$ we have $P(Z=n)=0$ if $n$ is odd, showing that it has not Poisson-distribution.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 2 at 18:44









drhabdrhab

99.9k544130




99.9k544130












  • $begingroup$
    Thank you! got it.
    $endgroup$
    – Alan
    Jan 2 at 18:45










  • $begingroup$
    You are welcome.
    $endgroup$
    – drhab
    Jan 2 at 18:46


















  • $begingroup$
    Thank you! got it.
    $endgroup$
    – Alan
    Jan 2 at 18:45










  • $begingroup$
    You are welcome.
    $endgroup$
    – drhab
    Jan 2 at 18:46
















$begingroup$
Thank you! got it.
$endgroup$
– Alan
Jan 2 at 18:45




$begingroup$
Thank you! got it.
$endgroup$
– Alan
Jan 2 at 18:45












$begingroup$
You are welcome.
$endgroup$
– drhab
Jan 2 at 18:46




$begingroup$
You are welcome.
$endgroup$
– drhab
Jan 2 at 18:46


















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