Find the distribution of $X$ given that $P(X=k)=\frac 23 (k+1)P(X=k+1)$

Question

A discrete random variable $X$ of values in $\mathbb N$ verifies the property that $$P(X=k)=\cfrac 23 (k+1)P(X=k+1)$$ What is the distribution of $X$?

I found that $$P(X\ge 0)=\sum_{k=0}^\infty P(X=k)=\sum_{k=0}^\infty\cfrac 23 (k+1)P(X=k+1)=\cfrac 23\sum_{k=1}^\infty kP(X=k)=\cfrac 23\text E(X)=1$$ $\ \ \ \ \ \ \ \ \text E(X) = 1.5$

I also found that $$P(X=k)=\cfrac{3^k}{2^k\cdot k!}\cdot P(X=0)$$ That is the only thing I could get out of the given property, I couldn't find the expression for $P(X=k)$ which is the actual question.

Answer 1

Very good! Note that $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots.$$ Our probabilities must add up to $1$. When we add up, we get $e^{3/2}\Pr(X=0)$. Now we can compute $\Pr(X=0)$, and therefore everything.

Answer 2

Sum these probabilities to get

$$\sum_{k=0}^{\infty}P(X=k)=1 $$ You can then solve for $P(X=0)$