I have a discrete random variable $X$ with range natural numbers including $0$. It's given the probability mass function of $X$ fullfills this recursion equation:
$$f_X(x) = \frac3x f(x − 1),\ x \in\mathbb N. $$
Now the question is what kind of a distribution is this and I have the solution that it is a Poisson distribution with
$$f_X(x) = \frac{e^{-3}3^x}{x!},\ x\in\mathbb N_0$$
I dont understand how? Can anyone please explain this to me?
This is a first-order difference equation, for which the techniques of solution are well known. If you multiply the equation through by $\ \frac{x!}{3^x}\ $, you get $$ \frac{x!\,f_X(x)}{3^x}=\frac{(x-1)!\,f_X(x-1)}{3^{x-1}}\ , $$ or, equivalently, $$ \frac{x!\,f_X(x)}{3^x}-\frac{(x-1)!\,f_X(x-1)}{3^{x-1}}=0\ . $$ Now summing this last equation from $\ x=1\ $ to $\ x=y\ $ gives \begin{align} 0&=\sum_{x=1}^y\left(\frac{x!\,f_X(x)}{3^x}-\frac{(x-1)!\,f_X(x-1)}{3^{x-1}}\right)\\ &=\sum_{x=1}^y\frac{x!\,f_X(x)}{3^x}-\sum_{x=1}^y\frac{(x-1)!\,f_X(x-1)}{3^{x-1}}\\ &=\sum_{x=1}^y\frac{x!\,f_X(x)}{3^x}-\sum_{x=0}^{y-1}\frac{x!\,f_X(x)}{3^x}\\ &=\frac{y!\,f_X(y)}{3^y}-f_X(0)\ \end{align} from which we conclude that $$ f_X(y)=\frac{3^y\,f_X(0)}{y!}\ . $$ Since $\ f_X(y)\ $ is a probability mass function, its values must sum to $1$: \begin{align} 1&=\sum_{y=0}^\infty f_X(y)\\ &=\sum_{y=0}^\infty\frac{3^y\,f_X(0)}{y!}\\ &=e^3f_X(0)\, \end{align} which gives $\ f_X(0)=e^{-3}\ $.