Let XGeometric13 and let YlX5l Find the PMF of YSolutionHere

Let X~Geometric(1/3), and let Y=lX-5l. Find the PMF of Y.

Here is some work to get you started:

Since XX is geometric with parameter 1313 we have

P(X=k)=(2/3)k1(1/3)P(X=k)=(2/3)k1(1/3)

where kNkN.

We want to find P(Y=k)P(Y=k), for kNkN, the probability mass function for YY.

P(Y=k)=P(k=|X5|)P(Y=k)=P(k=|X5|)

Now we use the definition of absolute value

=P(k=X5 or k=X5)=P(k=X5 or k=X5)

We then use rules of probability

=P(X=5k)+P(X=5+k)P(X=5k and X=5=k)=P(X=5k)+P(X=5+k)P(X=5k and X=5=k)

The 3rd term in the above sum is clearly zero (You could have noticed that the events were disjoint and then neglected to even write it)

=P(X=5k)+P(X=5+k)