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So I have a mathproblem where the Poisson distribution of the number of flowers X in one area is given by $\lambda_1=4$. The Poisson distribution for the numbers of flowers Y in the other area is given by $\lambda_2=6$. I am then asked to find an expression for $P(X=Y)$, and explain what condition that must be satisfied to calculate this. I'm not that good at probability problems, so what does $P(X=Y)$ mean, and how should I find this expression? I know that

$P(X)=\frac{4^X}{X!}e^{-4}$ and $P(Y)=\frac{6^Y}{Y!}e^{-6}$

Gary
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Mathomat55
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1 Answers1

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You can use $P(X=Y)=P(\bigcup_n \{ X= n\} \cap \{Y=n\})=\sum_nP(\{X=n\}\cap\{Y=n\})$. Now all you need is a condition to compute $P(\{X=n\}\cap\{Y=n\})$. The easiest would be independence.

Edit: Regarding your notation. As $X$ is a random variable and $P$ a probability measure, $\{X=n\}$ denotes the set of all $\omega\in\Omega$ such that $X(\omega)=n$. So writing $P(X)$ is abuse of notation, you should better write $P(X=n)=\frac{4^n}{n!}e^{-4}$. By $P(X=Y)$ we are looking for the probability of the set $\{\omega\in\Omega: X(\omega)=Y(\omega)\}$ which are indeed those $\omega$, such that there exists $n$ with $X(\omega)=n$ and $Y(\omega)=n$ which leads to the probability above.

  • Thank you for your answer :) So the condition for computing $P(X=Y)$ is that X and Y must be independent variables? Why so? And I don't know if I understand the equation $P(X=Y)=\sum_nP({X=n}\cap{Y=n})$. So if two events X and Y are independent $P(X\capY)=P(X)P(Y)$. Is the answer therefor equal to $P(X=Y)=\sum_n P(X=n)P(Y=n)$. Or I'm I wrong? – Mathomat55 Sep 14 '21 at 14:58
  • This would be one condition, as the comment under your post says, knowing the joint distribution is necessary. –  Sep 14 '21 at 15:01
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    Indeed, that’s what you obtain if you assume independence. –  Sep 14 '21 at 15:01
  • Okay, I see. Thank you both for you help! – Mathomat55 Sep 14 '21 at 15:04