Question

Let \(X\) have a pmf \(p(x)=\frac{1}{3}, x=1,2,3\), zero elsewhere. Find the pmf of \(Y=2 X+1\)

Short answer

The pmf for \(Y\) is \(p(y) = 1/3\) for \(y = 3, 5, 7\) and zero elsewhere.

Step-by-step solution

Determine the Possible Values of Y

The set of possible values that \(Y\) can take is based on the range of \(X\) and the relationship between \(X\) and \(Y\) in the equation \(Y=2X+1\). Here, \(X\) can take on 1, 2, or 3 as values. Therefore, \(Y\), which is equal to \(2X+1\), can take on the values 3, 5 or 7, when \(X\) is respectively 1, 2 or 3.

Transform the Probability Mass Function for Y

The next thing to do is change the pmf of \(X\) to derive the pmf of \(Y\). Since \(Y = 2X + 1\), substituting \(X\) values into the equation will provide the outcomes for \(Y\). As stated in the exercise, the pmf \(p(x) = 1/3\) for \(x = 1, 2, 3\). Since these \(X\) values transform to \(Y\) values 3, 5 and 7 respectively, the pmf \(p(y)\) for \(Y\) will also be 1/3 at \(y = 3, 5, 7\).

Verify the Result

Finally, to verify the result, confirm that the sum of the probabilities for all possible outcomes of \(Y\) is equal to 1. In this case, \(p(3) + p(5) + p(7) = 1/3 + 1/3 + 1/3 = 1\). Therefore, the solution is confirmed correct.