Question

Let \(X\) have the \(\operatorname{pmf} p(x)=\left(\frac{1}{2}\right)^{x}, x=1,2,3, \ldots\), zero elsewhere. Find the pmf of \(Y=X^{3}\).

Short answer

The pmf for \(Y\) is \(p(y)=(1/2)^(y^(1/3)), for y = 1, 8, 27, ...\), and zero elsewhere.

Step-by-step solution

Identify the Values of Y

Identify the possible values of \(Y\). Since \(Y = X^3\) and \(X\) takes only positive integer values, \(Y\) will only accept positive cubic integer values. Therefore, the possible values for \(Y\) are \(1^3, 2^3, 3^3, ....\). Thus, Y can be any cubic integer values like 1, 8, 27 and so on.

Calculate the Probabilities

Compute the probability that \(X\) acquires certain values which would give the identified values for \(Y\). Since \(Y = X^3\), the values that \(X\) would take to get certain \(Y-'s\) are those for which \(x^3 = y\). Hence, when you find the cubic root of a value of \(Y\), you will get the corresponding value for \(X\) that can give such a \(Y\) value. The provided pmf is \(p(x)=(1/2)^x\). So, the probability of \(Y = y\) would be \(p(y) = (1/2)^(cubic root of y)\).

Formulate the pmf for Y

Construct the probability mass function for \(Y\). Since \(Y = X^3\) and \(X\) accepts positive integer values only, the pmf for \(Y\) is not defined for \(y\) values which cannot be expressed as a perfect cube. Therefore, the pmf for \(Y\) is \( p(y)=(1/2)^(y^(1/3)), for y = 1, 8, 27, ...\), and zero elsewhere.