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S2 poisson distribution question

Two random variables X and Y have independent poisson distributions given by X~Poisson(1.4) and Y~Poisson(3.6) respectively.
Using the distributions of X and Y only, calculate P(X+Y =2).

I tried working it out doing this:
P(X+Y =2) = P(X=2) * P(Y=2)
0.241 * 0.177 = 0.042

However the answer in the book is 0.084 which is double what I got. Where have I gone wrong?
Original post by IDontKnowReally
Two random variables X and Y have independent poisson distributions given by X~Poisson(1.4) and Y~Poisson(3.6) respectively.
Using the distributions of X and Y only, calculate P(X+Y =2).

I tried working it out doing this:
P(X+Y =2) = P(X=2) * P(Y=2)
0.241 * 0.177 = 0.042

However the answer in the book is 0.084 which is double what I got. Where have I gone wrong?


But when x =2 and y= 2, x+y = 4.

Think about the different cases of x and y. As a starting point, it is immediate that x or y should each never be 3 or greater, if you can see why from the question.
Reply 2
Because they're independent, you can just add their means and then find that probability that X=2 of that mean.

[I'm also still learning this, so correct me if I'm wrong].
Original post by Synmatic
Because they're independent, you can just add their means and then find that probability that X=2 of that mean.

[I'm also still learning this, so correct me if I'm wrong].


Question does say "Using the distributions of X and Y only".

That aside, you are correct in that you can add their parameters.
Original post by SeanFM
But when x =2 and y= 2, x+y = 4.

Think about the different cases of x and y. As a starting point, it is immediate that x or y should each never be 3 or greater, if you can see why from the question.


How can I find what x and y are? Is it to do with the ratio of their means?
Original post by IDontKnowReally
How can I find what x and y are? Is it to do with the ratio of their means?


No, as above you can combine them using a property that is not required or used at A-level, and the question states that you cannot do this anyway. (I only mention this because you mentioned ratio, it's not quite ratio but the sum of the means eg if X~Po(10) and Y~Po(20) then X+Y~Po(30) but I repeat that this is not something you should know and the question says you are not allowed to use this method)


You tell me, what possible x and y values can you have in the equation x + y = 2?
(edited 7 years ago)
Original post by SeanFM
No, as above you can combine them using a property that is not required or used at A_level, and the question states that you cannot do this anyway.

You tell me, what possible x and y values can you have in the equation x + y = 2?


Anything greater than 0 and anything less than 2?
Original post by IDontKnowReally
Anything greater than 0 and anything less than 2?


That is right, but you just have to be careful. It can also include 2.

Can you now see how to solve the question? :h:
Original post by SeanFM
That is right, but you just have to be careful. It can also include 2.

Can you now see how to solve the question? :h:


No, im still a little confused, sorry.
Would it just be P(X<=2) * P(Y<=2) ?
Reply 9
I would just add the 2 distributions so you would let Z = X + Y so then Z~Po(5)
Then use your method to find the P(Z=2), it should give the answer the book gives.

Also, your method could be right (I'm not sure) as there are 2 ways in which X and Y can happen, X first then Y or Y first then X. You only did it one of those ways which would explain your answer being half the answer in the book.
Original post by IDontKnowReally
No, im still a little confused, sorry.
Would it just be P(X<=2) * P(Y<=2) ?


No, because in that probability if I selected, for example, X=2 and Y = 2 (which are in each of those probabilities) then X+Y = 4, which does not satisfy X+Y = 2, so that is a hint for you :smile:

So what are all the x and y values greater than or equal to 0 such that x + y = 2? And why is this useful for the question?
Original post by SeanFM
No, because in that probability if I selected, for example, X=2 and Y = 2 (which are in each of those probabilities) then X+Y = 4, which does not satisfy X+Y = 2, so that is a hint for you :smile:

So what are all the x and y values greater than or equal to 0 such that x + y = 2? And why is this useful for the question?


x=1, y=1
x=2. y=0
x=0, y=2
Could you do P(X=1) * P(Y=1) + P(x=2)* P(Y=0) + P(X=0) * P(Y=2) ?
Original post by IDontKnowReally
x=1, y=1
x=2. y=0
x=0, y=2
Could you do P(X=1) * P(Y=1) + P(x=2)* P(Y=0) + P(X=0) * P(Y=2) ?


Well done :h: with just very small nudges in the right direction you have worked it out for yourself.
Original post by SeanFM
Well done :h: with just very small nudges in the right direction you have worked it out for yourself.


Thanks for all your help! :smile:

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