Let $x_1,x_2,\dots,x_n$ be a set of independent and identically distributed random variables with distribution Poisson with parameter λ. Write the joint distribution of all those random variables. Simplify as much as possible your final answer and show work.
So I think that the joint probability of independent random variables is the product of all individual probability distribution function, but I don't actually understand how to implement that in this case, since it's for $n$ variables.
On
We just have a product of $n$ cumulative distribution functions, i.e., $$ P(X_1\le x_1,\ldots,X_n\le x_n)=P(X_1\le x_1)\cdot\ldots\cdot P(X_n\le x_n)=e^{-\lambda}\prod_{i=1}^n\sum_{j=0}^{\lfloor x_i\rfloor}\frac{\lambda^j}{j!} $$ for any $x_1,\ldots,x_n\in\mathbb R$, where $\lfloor\cdot\rfloor$ is the floor function. I am not sure if it is possible to simplify this expression any further.
The pdf of the Poisson distribution is
$$p(k) = \frac{\lambda ^k}{k!} \exp(-\lambda)$$
The joint pdf is the product of the pdfs of all $n$ independent variables $x_i$. Which is given by
$$p_n = p(x_1,x_2,...,x_n) =\prod _{i=1}^n p(x_i) $$
which can be simplified to
$$p_n = \lambda^{\sum _{i=1}^n x_i}\exp(- n \lambda)\; \prod_{i=1}^n \frac{1}{x_i !}$$