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If $\text P(X>k+1)=\cfrac 12 \text P(X>k)$, $k\in \mathbb N^*$ what is the distribution of $X$?

So I did; $$\text P(X=k+1)= \text P(X>k)- \text P(X>k+1)=\cfrac 12 \text P(X>k)$$ $$\text P(X=k)=\cfrac 12 \text P(X>k-1)$$ I don't know what to deduce from this.

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Does $k\in \mathbb{Z}^+$? and is $X$ discrete or continuous? – user17762 Nov 7 '12 at 18:50
@Marvis $k\in \mathbb N^*$ – user31280 Nov 7 '12 at 18:52
Is $X$ discrete or continuous? And is $X>0$? – user17762 Nov 7 '12 at 18:57
@Marvis the question didn't specify? – user31280 Nov 7 '12 at 18:59

I will rewrite the last equation that you got as: $P(X=k)=0.5[P(X=k)+P(X=k+1)+P(X=k+2)+....]$....(1) Thus: $P(X=k+1)=0.5[P(X=k+1)+P(X=k+2)+P(X=k+3)+...]$...(2), if we subtract (2) from (1),we get: $P(X=k+1)-P(X=k)=-0.5 P(X=k)$....(3) Since $P(X=1)+P(X=2)+...=1$, therefore recurrence relation (3) can be solved easily.

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up vote 1 down vote accepted

Assuming $k \in \mathbb{N}$, and assuming that $X>0$, we get that $P(X>0) = 1$. This gives us that $P(X>1) = \dfrac12$ and in general $P(X> k ) = \dfrac1{2^k}$. Hence, $$P(X \in (k-1,k]) = P(X>k-1) - P(X>k) = \dfrac1{2^{k-1}} - \dfrac1{2^k} = \dfrac1{2^k}$$ If $X$ is a discrete random variable taking only values in $\mathbb{Z}^+$, then $$P(X \in (k-1,k]) = P(X = k) = \dfrac1{2^k}$$

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