Small Poisson Probability Question | Stochastic Vector with Discrete Support

[Hartogsohn26]

Hello

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 $$p_{T,U}$$ given by.

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

where $$\lambda > 0$$

1)

describe the support for $$\mathrm{supp} (P_{T,U})$$

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

2)

show that the probability function $$p_T$$ and $$p_U$$ for T and U is.

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

and

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

which inturn means $$U \sim po(\lambda)$$

Solution (2)

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


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


Solution is $$P(T = U) = P(t \cup u)$$?


Best Regards
Hartogsohn

 

[lippyka]

Solution (3) P(T=U) = \sum_{t=0}^{\infty}\sum_{u=0}^{\infty} P(T=t, U=u) = \sum_{t=0}^{\infty} \frac{1}{3} e^{-1} \frac{1^u}{u!} = \frac{2}{3} e^{-1}.