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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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Be careful with the first term in the series. If $X$ assumes values in $\mathbb{N}$, the first term is $P(X=1)$, not $P(X=0)$. –  user1551 Nov 7 '12 at 20:31
    
Depends on what definition one has of $\N$ –  Jean-Sébastien Nov 7 '12 at 20:33
    
@user1551 and why is that? is $0\notin \mathbb N$? –  user31280 Nov 7 '12 at 20:34
    
@F'OlaYinka it depends on how you define it. some author will include it, some others won't –  Jean-Sébastien Nov 7 '12 at 20:40
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If you want to start at $n=1$, that will not change things very much. About worries about not seeing something "simple like that" you identified the probability part of the calculation correctly. If you didn't know a simple expression for the sum, you could call it $C$ and say that $\Pr(X=k)=\frac{1}{C}$ times stuff you can write down. That would get you most of the marks for the question. –  André Nicolas Nov 7 '12 at 20:45
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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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you mean $Pr(X=0)$ no? –  Jean-Sébastien Nov 7 '12 at 20:31
    
Yes, changed. ${}{}{}$ –  André Nicolas Nov 7 '12 at 20:32
    
My exam is in two days and i can't see something as obvious as this, I pity myself. Thank you very much! –  user31280 Nov 7 '12 at 20:34
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Sum these probabilities to get

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

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