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Let $\alpha$ be the probability that a geometric random variable $X$ with parameter p is an even number

a) Find $\alpha$ using the identity $\alpha=\sum_{i=1}^{\infty}P[X=2i]$

b)Find $\alpha$ by conditioning on wether $X=1$ or $X>1$

My attempt for a):

$$\alpha=\sum_{i=1}^{\infty}P[X=2i]=\sum_{i=1}^{\infty}p(1-p)^{2i-1}=p(1-p)\sum_{i=1}^{\infty}(1-p)^{2(i-1)}$$ Letting $j=i-1$

$$p(1-p)\sum_{i=1}^{\infty}(1-p)^{2(i-1)}=p(1-p)\sum_{j=0}^{\infty}(1-p)^{2j}={{p(1-p)}\over {1-(1-p)^2}}$$

hence $$\alpha={{1-p}\over {2-p}}$$

The thing is that I´m having trouble for b): Let $E$ be the event that a geometric random variable $X$ with parameter p is an even number, so : $P(E)=\alpha$ I need to find this probability by conditioning on wether $X=1$ or $X>1$ therefore:

$$P(E)=P(E|X=1)P(X=1)+P(E|X>1)P(X>1)$$

but $P(E|X=1)=0$ hence $P(E)=P(E|X>1)P(X>1)$ then we have that $$P(X>1)=1-P(X\le 1)=1-p$$

But I´m having trouble computing $P(E|X>1)$ Can you help me please? I would really appreciate it :)

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  • $\begingroup$ Hint: $\mathsf P(E\mid X>1) = 1-\mathsf P(E)$, The probability of the rv being even given it is greater than one is equal to the probability of the rv being odd. $\endgroup$ – Graham Kemp Nov 10 '14 at 0:46
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We have $$\eqalign{ P(\hbox{$X$ is even}) &=P(\hbox{$X$ is even}\mid X=1)P(X=1)+P(\hbox{$X$ is even}\mid X>1)P(X>1)\cr &=P(\hbox{$X$ is even}\mid X>1)P(X>1)\cr}$$ since obviously the first term on the RHS is zero. Now the probability that $X$ is even, given that the first trial was a failure, is the same as the probability that $X$ is odd (because after the first failure you have an identical experiment, but you need an odd result in this experiment in order to get an even result overall). So $$\alpha=(1-\alpha)(1-p)$$ and solving gives $$\alpha=\frac{1-p}{2-p}\ .$$

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Let $\alpha$ be the required probability. In order for $X$ to be even, we must have a failure on the first trial, and have the number of trials after the first be odd. The probability of that, given that the first trial was a failure, is $1-\alpha$. Thus $$\alpha=(1-p)(1-\alpha).$$ Solve for $\alpha$.

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  • $\begingroup$ I really appreciate your time, thanks @André :) $\endgroup$ – user128422 Nov 10 '14 at 4:17
  • $\begingroup$ You are welcome. Conditioning on the first outcome (or the first few) can be very useful both for calculations of probability and for calculations of expectation. $\endgroup$ – André Nicolas Nov 10 '14 at 6:31

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