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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 $alpha=sum_i=1^inftyP[X=2i]$

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

My attempt for a):

$$alpha=sum_i=1^inftyP[X=2i]=sum_i=1^inftyp(1-p)^2i-1=p(1-p)sum_i=1^infty(1-p)^2(i-1)$$ 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 $$alpha=1-pover 2-p$$

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 $$P(X>1)=1-P(Xle 1)=1-p$$

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

We have
$$eqalign
P(hbox$X$ is even)
&=P(hbox$X$ is evenmid X=1)P(X=1)+P(hbox$X$ is evenmid X>1)P(X>1)cr
&=P(hbox$X$ is evenmid X>1)P(X>1)cr$$
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
$$alpha=frac1-p2-p .$$

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 $alpha$.