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Your assumptions quickly leads to contradictions except for certain values of $\lambda_1,\lambda_2,a$ and $b$. Suppose we know the joint probability mass function in $(x_1,x_2)=(1,1)$, that is, $p(1,1)$. Keeping in mind that the joint probability mass function and the conditional distributions are proportional, we can now derive an expression for $p(0,0)$ in terms of $p(1,1)$ in two ways. From the assumed conditional Poisson distributions it follows that \begin{equation} \frac{p(1,0)}{p(1,1)} = \frac{p_{X_2|X_1=1}(0)}{p_{X_2|X_1=1}(1)} = \frac{e^{-\lambda_2+b}}{e^{-\lambda_2+b}(\lambda_2+b)} = \frac1{\lambda_2+b}. \end{equation} Similarly, \begin{equation} \frac{p(0,0)}{p(1,0)} = \frac{p_{X_1|X_2=0}(0)}{p_{X_1|X_2=0}(1)} = \frac{e^{-\lambda_1}}{e^{-\lambda_1}\lambda_1} = \frac1{\lambda_1}. \end{equation} Hence, \begin{equation} p(0,0) = \frac1{\lambda_1(\lambda_2+b)}p(1,1). \end{equation} Doing the same argument but going via $p(0,1)$ leads to \begin{equation} p(0,0) = \frac1{\lambda_2(\lambda_1+a)}p(1,1). \end{equation} At least if $\lambda_1 b\neq\lambda_2 a$, the two last equations can't both be true and your assumptions are inconsistent.

Similar arguments leads to additional constraints on $\lambda_1,\lambda_2,a$ and $b$ so I suspect that the only possible consistent case is $a=b=0$ in which case $X_1$ and $X_2$ are independent.

Your assumptions quickly leads to contradictions except for certain values of $a$ and $b$. Keeping in mind that the joint probability mass function and the conditional distributions are proportional, we can derive an expression for $p(x_1,x_2)$ in terms of $p(0,0)$ in two ways. From the assumed conditional Poisson distributions it follows that \begin{equation} \frac{p(0,x_2)}{p(0,0)} = \frac{p_{X_2|X_1=0}(x_2)}{p_{X_2|X_1=0}(0)} = \frac{e^{-\lambda_2}\lambda_2^{x_2}/x_2!}{e^{-\lambda_2}} = \lambda_2^{x_2}/x_2!. \end{equation} Similarly, \begin{equation} \frac{p(x_1,x_2)}{p(0,x_2)} = \frac{p_{X_1|X_2=x_2}(x_1)}{p_{X_1|X_2=x_2}(0)} = \frac{e^{-(\lambda_1+a x_2)}(\lambda_1+a x_2)^{x_1}/x_1!}{e^{-(\lambda_1+a x_2)}} = (\lambda_1+a x_2)^{x_1}/x_1! \end{equation} Hence, \begin{equation} p(x_1,x_2) = \frac{(\lambda_1+a x_2)^{x_1}\lambda_2^{x_2}}{x_1!x_2!}p(0,0). \end{equation} Doing the same argument but going via $p(x_1,0)$ leads to \begin{equation} p(x_1,x_2) = \frac{\lambda_1^{x_1}(\lambda_2+b x_1)^{x_2}}{x_1!x_2!}p(0,0). \end{equation} The two last equations can both be true only if \begin{equation} (\lambda_1+a x_2)^{x_1}\lambda_2^{x_2} = \lambda_1^{x_1}(\lambda_2+b x_1)^{x_2} \end{equation} for all $(x_1,x_2)$ which is only possible if $a=b=0$, that is, if $X_1$ and $X_2$ are independent. Otherwise, the assumption that the conditional distributions are Poisson lead to a contradiction and are thus inconsistent.

Your assumptions quickly leads to contradictions except for certain values of $\lambda_1,\lambda_2,a$ and $b$. Suppose we know the joint probability mass function in $(x_1,x_2)=(1,1)$, that is, $p(1,1)$. Keeping in mind that the joint probability mass function and the conditional distributions are proportional, we can now derive an expression for $p(0,0)$ in terms of $p(1,1)$ in two ways. From the assumed conditional Poisson distributions it follows that \begin{equation} \frac{p(1,0)}{p(1,1)} = \frac{p_{X_2|X_1=1}(0)}{p_{X_2|X_1=1}(1)} = \frac{e^{-\lambda_2+b}}{e^{-\lambda_2+b}(\lambda_2+b)} = \frac1{\lambda_2+b}. \end{equation} Similarly, \begin{equation} \frac{p(0,0)}{p(1,0)} = \frac{p_{X_1|X_2=0}(0)}{p_{X_1|X_2=0}(1)} = \frac{e^{-\lambda_1}}{e^{-\lambda_1}\lambda_1} = \frac1{\lambda_1}. \end{equation} Hence, \begin{equation} p(0,0) = \frac1{\lambda_1(\lambda_2+b)}p(1,1). \end{equation} Doing the same argument but going via $p(0,1)$ leads to \begin{equation} p(0,0) = \frac1{\lambda_2(\lambda_1+a)}p(1,1). \end{equation} At least if $\lambda_1(\lambda_2+b)\neq\lambda_2(\lambda_1 +a)$, the two last equations can't both be true and your assumptions are inconsistent.

Similar arguments leads to additional constraints on $\lambda_1,\lambda_2,a$ and $b$ so I suspect that the only possible consistent case is $a=b=0$ in which case $X_1$ and $X_2$ are independent.

Your assumptions quickly leads to contradictions except for certain values of $\lambda_1,\lambda_2,a$ and $b$. Suppose we know the joint probability mass function in $(x_1,x_2)=(1,1)$, that is, $p(1,1)$. Keeping in mind that the joint probability mass function and the conditional distributions are proportional, we can now derive an expression for $p(0,0)$ in terms of $p(1,1)$ in two ways. From the assumed conditional Poisson distributions it follows that \begin{equation} \frac{p(1,0)}{p(1,1)} = \frac{p_{X_2|X_1=1}(0)}{p_{X_2|X_1=1}(1)} = \frac{e^{-\lambda_2+b}}{e^{-\lambda_2+b}(\lambda_2+b)} = \frac1{\lambda_2+b}. \end{equation} Similarly, \begin{equation} \frac{p(0,0)}{p(1,0)} = \frac{p_{X_1|X_2=0}(0)}{p_{X_1|X_2=0}(1)} = \frac{e^{-\lambda_1}}{e^{-\lambda_1}\lambda_1} = \frac1{\lambda_1}. \end{equation} Hence, \begin{equation} p(0,0) = \frac1{\lambda_1(\lambda_2+b)}p(1,1). \end{equation} Doing the same argument but going via $p(0,1)$ leads to \begin{equation} p(0,0) = \frac1{\lambda_2(\lambda_1+a)}p(1,1). \end{equation} At least if $\lambda_1 b\neq\lambda_2 a$, the two last equations can't both be true and your assumptions are inconsistent.

Similar arguments leads to additional constraints on $\lambda_1,\lambda_2,a$ and $b$ so I suspect that the only possible consistent case is $a=b=0$ in which case $X_1$ and $X_2$ are independent.

Your assumptions quickly leads to contradictions except for certain values of $a$ and $b$. Suppose we know the joint probability mass function in $(x_1,x_2)=(1,1)$, that is, $p(1,1)$. Keeping in mind that the joint probability mass function and the conditional distributions are proportional, we can now derive an expression for $p(0,0)$ in terms of $p(1,1)$ in two ways. From the assumed conditional Poisson distributions it follows that \begin{equation} \frac{p(1,0)}{p(1,1)} = \frac{p_{X_2|X_1=1}(0)}{p_{X_2|X_1=1}(1)} = \frac{e^{-\lambda_2+b}}{e^{-\lambda_2+b}(\lambda_2+b)} = \frac1{\lambda_2+b}. \end{equation} Similarly, \begin{equation} \frac{p(0,0)}{p(1,0)} = \frac{p_{X_1|X_2=0}(0)}{p_{X_1|X_2=0}(1)} = \frac{e^{-\lambda_1}}{e^{-\lambda_1}\lambda_1} = \frac1{\lambda_1}. \end{equation} Hence, \begin{equation} p(0,0) = \frac1{\lambda_1(\lambda_2+b)}p(1,1). \end{equation} Doing the same argument but going via $p(0,1)$ leads to \begin{equation} p(0,0) = \frac1{\lambda_2(\lambda_1+a)}p(1,1). \end{equation} At least if $\lambda_1(\lambda_2+b)\neq\lambda_2(\lambda_1 +a)$, the two last equations can't both be true and your assumptions are inconsistent.

Similar arguments leads to additional constraints on $\lambda_1,\lambda_2,a$ and $b$ so I suspect that the only possible consistent case is $a=b=0$ in which case $X_1$ and $X_2$ are independent.

Your assumptions quickly leads to contradictions except for certain values of $\lambda_1,\lambda_2,a$ and $b$. Suppose we know the joint probability mass function in $(x_1,x_2)=(1,1)$, that is, $p(1,1)$. Keeping in mind that the joint probability mass function and the conditional distributions are proportional, we can now derive an expression for $p(0,0)$ in terms of $p(1,1)$ in two ways. From the assumed conditional Poisson distributions it follows that \begin{equation} \frac{p(1,0)}{p(1,1)} = \frac{p_{X_2|X_1=1}(0)}{p_{X_2|X_1=1}(1)} = \frac{e^{-\lambda_2+b}}{e^{-\lambda_2+b}(\lambda_2+b)} = \frac1{\lambda_2+b}. \end{equation} Similarly, \begin{equation} \frac{p(0,0)}{p(1,0)} = \frac{p_{X_1|X_2=0}(0)}{p_{X_1|X_2=0}(1)} = \frac{e^{-\lambda_1}}{e^{-\lambda_1}\lambda_1} = \frac1{\lambda_1}. \end{equation} Hence, \begin{equation} p(0,0) = \frac1{\lambda_1(\lambda_2+b)}p(1,1). \end{equation} Doing the same argument but going via $p(0,1)$ leads to \begin{equation} p(0,0) = \frac1{\lambda_2(\lambda_1+a)}p(1,1). \end{equation} At least if $\lambda_1(\lambda_2+b)\neq\lambda_2(\lambda_1 +a)$, the two last equations can't both be true and your assumptions are inconsistent.

Similar arguments leads to additional constraints on $\lambda_1,\lambda_2,a$ and $b$ so I suspect that the only possible consistent case is $a=b=0$ in which case $X_1$ and $X_2$ are independent.