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I just found this forum, was recommended it by a person that I know and have a small probability question regarding Poisson distribution.

Let (T,U) be a to dimensional discrete stochastic vector with the probability function [tex]p_{T,U} [/tex] given by.

[tex]P(T=t, U = u) = \left{ \begin{array}{cccc} \frac{1}{3} \cdot e^{-\lambda} \frac{\lambda^u}{u!} & u \in \{-1,0,1\} & \mathrm{and} & t \in \{0, 1, \ldots\} \\0 & \mathrm{elsewhere.} \end{array} [/tex]

where [tex]\lambda > 0 [/tex]

1)

describe the support for [tex]\mathrm{supp} (P_{T,U}) [/tex]

Solution (1) the support [tex]\mathrm{supp} (P_{T,U}) [/tex] is the set of values for the real valued probability function P which produces non-negative values. therefore [tex]\mathrm{supp} (P_{T,U}) = \{0,1, \ldots \} [/tex]

2)

show that the probability function [tex]p_T [/tex] and [tex]p_U [/tex] for T and U is.

[tex]P(T = t) = P_{T} = \left{ \begin{array}{cccc} \frac{1}{3} & t \in \{-1,0,1\} \\0 & \mathrm{elsewhere.} \end{array} [/tex]

and

[tex]P(U = u) = P_{U} = \left{ \begin{array}{cccc} e^{-\lambda} \frac{\lambda^{u}}{u!} & u \in \{0,1,\ldots\} \\0 & \mathrm{elsewhere.} \end{array} [/tex]

which inturn means [tex]U \sim po(\lambda) [/tex]

Solution (2)

How do I show this ? If not as above. Or do I show that they have same variance??

(3) Assume that [tex]\lambda = 1 [/tex] then [tex]P(T=U) = \frac{2}{3} e^{-\lambda} [/tex]

Solution is [tex]P(T = U) = P(t \cup u)[/tex]???

Best Regards

Hartogsohn

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# Small Poisson Question

Can you offer guidance or do you also need help?

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