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.

Levi Rasmussen

Levi Rasmussen

Answered question

2022-07-22

Find the probability that a geometric random variable X is an even number
Let α be the probability that a geometric random variable X with parameter p is an even number
a) Find α using the identity α = i = 1 P [ X = 2 i ].
b)Find α by conditioning on wether X = 1 or X > 1

Answer & Explanation

esbalatzaj

esbalatzaj

Beginner2022-07-23Added 15 answers

Step 1
We have P ( X  is even ) = P ( X  is even X = 1 ) P ( X = 1 ) + P ( X  is even X > 1 ) P ( X > 1 ) = P ( X  is even X > 1 ) P ( X > 1 ) 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).
Step 2
So α = ( 1 α ) ( 1 p ) and solving gives α = 1 p 2 p   .
skilpadw3

skilpadw3

Beginner2022-07-24Added 4 answers

Step 1
Let α 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 α.
Step 2
Thus α = ( 1 p ) ( 1 α ) .. Solve for α.

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