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Bivariate Poisson |
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| Apr21-07, 04:04 PM | #1 |
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Bivariate Poisson
I have 2 dependent random Poisson distributed variables, [tex]X[/tex] and [tex]Y[/tex]. I have that [tex]E[X] = mu[/tex] and [tex]E[Y] = c*mu[/tex] where [tex]c[/tex] is just a constant.
Now I'm trying to get the joint distribution of [tex]XY[/tex]. I've found the expression of the bivariate Poisson distribution but the problem is in order to use it I have to define [tex]X[/tex] and [tex]Y[/tex] as [tex]X = X' + Z[/tex] and [tex] Y = Y' + Z [/tex] where [tex]X', Y', Z'[/tex] are independent Poisson distributions with [tex]E[X'] = (mu - d)[/tex], [tex]E[Y'] = (c*mu - d)[/tex] and [tex]E[Z'] = d[/tex]. So basically my question is how do I get the parameter [tex]d[/tex]?? Is there any formal way to get it?? |
| Apr21-07, 04:16 PM | #2 |
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You have not been given enough information. X and Y could be independent or else Y=cX or something in between.
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| Apr21-07, 04:20 PM | #3 |
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Well, X and Y are definitley dependent, it is always [tex]E[Y] = cE[X][/tex].
Does that help?? If not, what more information is needed?? In the paper I have about these bivariate Poisson distribution it also states that [tex]P(X|Y) = d/(c*mu + d)[/tex] and also [tex]P(Y|X) = d/(mu + d)[/tex], if that's any help? |
| Apr22-07, 03:57 PM | #4 |
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Recognitions:
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Bivariate PoissonYour additional equation could be the key to the solution. |
| Apr22-07, 07:21 PM | #5 |
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| Apr22-07, 08:14 PM | #6 |
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Ummm, if P(Y|X) is a function that doesn't depend on X, then Y and X are independent.
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| Apr23-07, 12:48 PM | #7 |
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| Apr23-07, 05:31 PM | #8 |
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P(X = x and Y = y) = P(X = x) * P(Y = y).(Equivalently, P(X = x | Y = y) = P(X = x)) Two random variables are dependent if and only if they are not independent. |
| May1-09, 06:36 PM | #9 |
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Any idea to operate with Excel???
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