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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.

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    @F'Ola Yinka: During my undergraduate days, textbooks tended to denote $\{0,1,2,\ldots\}$ by $\mathbb{N}_0$. Without the "nought", $\mathbb{N}$ starts from 1. Maybe it's different today.2012-11-07

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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.

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    My exam is in two days and i can't see something as obvious as this, I pity myself. Thank you very much!2012-11-07
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Sum these probabilities to get

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