# Two Poisson distributed random variables

Homework Statement:
Have two Poisson distributed random variables, with parameter ##\lambda##=2
Relevant Equations:
probably
How do I evaluate
P(X-Y=0)=?

etotheipi
It's a little tricky. The difference of two Poisson variables is a Skellam distribution (as opposed to a sum of Poisson variables, which is just another Poisson variable with parameter ##\lambda_1 + \lambda_2##), so if you want you could look up the formulae for that.

You can try and work it out for yourself, however, by considering$$P(X = Y) = \sum_{k=0}^{\infty} P(X=k)P(Y=k) = \dots$$You will end up with an infinite sum for which you will require a modified Bessel function of the first kind, $$I_0(t) = \sum_{k=0}^{\infty} \frac{(\frac{1}{2}t)^{2k}}{(k!)^2}$$

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member 587159
member 587159
You can't say anything with the information given. For example, if ##X = Y## are such random variables then ##\Bbb{P}(X-Y = 0) = \Bbb{P}(X=Y) = 1## but in general it is possible that ##\Bbb{P}(X - Y=0) < 1##.

Probably, you want to ask about the case where ##X## and ##Y## are independent. Then as @etotheipi this follows a Skellam distribution.

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etotheipi
it was a question in a multiple choice question. i was trying to find out why that wasn't the correct answer:
it was stating that
##P \left( X-Y=0 \right)=1##

hence i was trying to compute it

etotheipi
it was a question in a multiple choice question. i was trying to find out why that wasn't the correct answer:
it was stating that
##P \left( X-Y=0 \right)=1##

hence i was trying to compute it

Do you have the verbatim problem statement?

FactChecker
FactChecker
Gold Member
Do you have the verbatim problem statement?
Sometimes it is so surprising to see what details are left out of the official problem statement.

etotheipi

member 587159
Yes, I agree with that answer.

etotheipi
Do you have the verbatim problem statement?
me too but, i want to know/understand why P[X-Y=]=1 is not correct.

etotheipi
me too but, i want to know/understand why P[X-Y=]=1 is not correct.

Well if the two continuous random variables are independent then it's definitely not true, since we could have, for instance ##X = 1## and ##Y = 33##. And a whole load of other combinations with ##X \neq Y##. Even if they are dependent, you will still generally have combinations with ##X \neq Y##.

If you impose a constraint that e.g. ##X =Y##, like @Math_QED explained nicely above, then you might well find that ##P(X-Y = 0) = 1##. But in general case that's not true.

member 587159 and DottZakapa
member 587159
Some other easy examples. The first is a discrete one, the second a continuous one.

Flip a fair coin. We have ##\Omega = \{H,T\}##, that is either we end up with heads or tails. The random variables ##I_{\{H\}}## (indicator function on the set ##\{H\}##) and ##I_{\{T\}}## are identical distributed yet they are unequal everywhere.

Let ##U## be a uniform distribution on ##[-1,1]##. Then ##-U## has the same distribution yet they are not identic.

etotheipi