Two Poisson distributed random variables

[DottZakapa]

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, $$I_0(t) = \sum_{k=0}^{\infty} \frac{(\frac{1}{2}t)^{2k}}{(k!)^2}$$

 

[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.

 

[DottZakapa]

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]

Do you have the verbatim problem statement?

Sometimes it is so surprising to see what details are left out of the official problem statement.

 

[DottZakapa]

 

[member 587159]

Yes, I agree with that answer.

 

[DottZakapa]

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]

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.