Section 2.6

Transformations

Question 2.6.1

Let \(X \sim {\mathrm{Unif}}[L,R]\). Let \(Y = cX + d\), where \(c > 0\). Prove that \(Y \sim {\mathrm{Unif}}[cL+d, cR+d]\).

Let \(h(x) = cx + d\). Then \(Y = h(X)\) and \(h\) is strictly increasing. Then \(h^{-1}(y) = (y - d)/c\) and \(\frac{d}{dy}h^{-1}(y) = \frac{1}{c}\)

so \[ f_Y(y) = f_X(h^{-1}(y)) \, / \, |h'(h^{-1}(y))| = f_X\left(\frac{y-d}{c}\right) \, / \, c \] which equals \(\dfrac{1}{c(R-L)}I_{[cL+d, cR+d]}(y)\). Hence, \(Y \sim {\mathrm{Unif}}[cL+d, cR+d]\).

Question 2.6.2

Let \(X \sim {\mathrm{Exp}}(\lambda)\). Let \(Y = X^3\). Compute the density \(f_Y\) of \(Y\).

Let \(h(x) = x^3\). Then \(Y = h(X)\) and \(h\) is strictly increasing, and \(h^{-1}(y) = y^{1/3}\). Hence, \[ f_Y(y) = f_X(h^{-1}(y)) \, / \, |h'(h^{-1}(y))| = f_X(y^{1/3}) \, / \, 3(y^{1/3})^2I_{(0, \infty)}(y) \] which equals \[ f_Y(y) = \frac{\lambda e^{-\lambda y^{1/3}}}{3y^{2/3}} = (\lambda/3) y^{-2/3} e^{-\lambda y^{1/3}}I_{(0, \infty)}(y) \]

Question 2.6.3

Let \(X \sim {\mathrm{Unif}}[0,3]\). Let \(Y = X^2\). Compute the density function \(f_Y\) of \(Y\).

Let \(h(x) = x^2\). Then \(Y = h(X)\), and \(h\) is strictly increasing over the region \(\{0 < x < 3\}\) where \(f_X(x) > 0\). Also, \(h^{-1}(y) = y^{1/2}\) on this region.

Hence, \(f_Y(y) = 0\) unless \(y > 0\) and \(0 < y^{1/2} < 3\), i.e., \(0 < y < 9\), in which case \[ \begin{aligned} f_Y(y) &= f_X(h^{-1}(y)) \, / \, |h'(h^{-1}(y))| \\ &= f_X(y^{1/2}) \, / \, 2y^{1/2} \\ &= \frac{1}{3} \cdot \frac{1}{2y^{1/2}} I_{(0, 9)}(y)\\ &= \frac{1}{6\sqrt{y}}I_{(0, 9)}(y) \end{aligned} \] for \(0 < y < 9\).

Question 2.6.3

Let \(X\) have density function \(f_X(x) = x^3/4\) for \(0 < x < 2\), otherwise \(f_X(x) = 0\).

  1. Let \(Y = X^2\). Compute the density function \(f_Y(y)\) for \(Y\).

  2. Let \(Z = \sqrt{X}\). Compute the density function \(f_Z(z)\) for \(Z\).

The inverse function of \(h(x) = x^2\) is \(h^{-1}(y) = y^{1/2}\), and the derivative of \(h\) is \(h'(x) = 2x\). Hence, \[ f_Y(y) = f_X(y^{1/2}) \, / \, |h'(y^{1/2})| = \frac{(y^{1/2})^3/4}{2y^{1/2}}I_{(0, 4)}(y) = \frac{y}{8}I_{(0, 4)}(y) \]

Since \(h(x) = x^{1/2}\), we have \(h^{-1}(z) = z^2\) and \(h'(x) = \tfrac{1}{2}x^{-1/2}\). So, \[ f_Z(z) = f_X(z^2) \, / \, |h'(z^2)| = f_X(z^2) \cdot 2zI_{(0, \sqrt(2))}(z) = \frac{(z^2)^3}{4} \cdot 2z = \frac{z^7}{2}I_{(0, \sqrt(2))}(z) \]

Question 2.6.4

Let \(X\) be the number showing when a fair six-sided die is rolled, so that \(X \in \{1,2,3,4,5,6\}\) with \(p_X(x) = 1/6\) for each \(x\). Let \[ Y = h(X) = (X-3)^2 \]

  1. What is the PMF of \(Y\)?

  2. Compute \(\mathbb{P}(Y \leq 4)\).

Since \(X\) is discrete, for each possible value \(y\) of \(Y\), \[ p_Y(y) = \sum_{x \,:\, h(x) = y} p_X(x) \]

First, compute \(h(x) = (x-3)^2\) for each \(x \in \{1,\dots,6\}\):

\(x\) 1 2 3 4 5 6
\(h(x)=(x-3)^2\) 4 1 0 1 4 9

Note that \(h\) is not one-to-one here — both \(x=2\) and \(x=4\) map to \(y=1\), and both \(x=1\) and \(x=5\) map to \(y=4\). So we must sum probabilities over all \(x\) values that map to the same \(y\):

\[ h^{-1}(0) = \{3\}, \quad h^{-1}(1) = \{2,4\}, \quad h^{-1}(4) = \{1,5\}, \quad h^{-1}(9) = \{6\} \]

Therefore, \[ p_Y(0) = p_X(3) = \frac{1}{6}, \qquad p_Y(1) = p_X(2) + p_X(4) = \frac{2}{6} = \frac{1}{3}, \] \[ p_Y(4) = p_X(1) + p_X(5) = \frac{2}{6} = \frac{1}{3}, \qquad p_Y(9) = p_X(6) = \frac{1}{6}, \] with \(p_Y(y) = 0\) for all other \(y\).

\[ \mathbb{P}(Y \leq 4) = p_Y(0) + p_Y(1) + p_Y(4) = \frac{1}{6} + \frac{1}{3} + \frac{1}{3} = \frac{5}{6} \]