Lecture #14: Change of Variables

Finding the Distribution of a Function of Random Variables

We’ve talked a bit about how to find the distribution of a function of random variables, for certain classes of functions, but let’s focus some more on it.

The Method of Moment-Generating Functions

There are two important properties of the MGF to remember:

1. If \(Y = aX\) where \(a \in \mathbb{R}\) is a constant, then: \[ M_Y(t) = M_X(at). \]

2. If \(Z = \dsum_{i=1}^n X_i\), then \[ M_Z(t) = \prod_{i=1}^n M_{X_i}(t). \]

The Method of Moment-Generating Functions

Summary of the Moment-Generating Function Method

Let \(U\) be a function of the random variables \(Y_1, Y_2, \dots, Y_n\).

1. Find the moment-generating function for \(U\), \(m_U(t)\).
2. Compare \(m_U(t)\) with other well-known moment-generating functions. If \(m_U(t)= m_V(t)\) for all values of \(t\), this implies that \(U\) and \(V\) have identical distributions.

Example: The Method of Moment-Generating Functions

Example: Suppose that \(X_i \overset{i.i.d.}{\sim} \text{Exponential}(\beta)\). If \(Y = \dsum_{i=1}^n X_i\), show that the distribution of \(Y\) is: \[ Y \sim \text{Gamma}(\alpha = n, \beta = \beta). \]

The Method of Distribution Functions

Example: A process for refining sugar yields up to \(1\) ton of pure sugar per day, but the actual amount produced, \(Y\), is a random variable because of machine breakdowns and other slowdowns. Suppose that \(Y\) has density function given by \[ f(y) = \begin{cases} 2y, & 0 \le y \le 1, \\ 0, & \text{otherwise.} \end{cases} \] The company is paid at the rate of \(\$300\) per ton for the refined sugar, but it also has a fixed overhead cost of \(\$100\) per day. Thus the daily profit, in hundreds of dollars, is \(U = 3Y− 1\). Find the probability density function for \(U\).

The Method of Distribution Functions

Summary of the Distribution Function Method

Let \(U\) be a function of the random variables \(Y_1, Y_2, \dots, Y_n\).

1. Find the region \(U = u\) in the \((y_1, y_2, \dots, y_n)\) space.
2. Find the region \(U \le u\).
3. Find \(F_U(u)= P(U \le u)\) by integrating \(f(y_1, y_2, \dots, y_n)\) over the region \(U \le u\).
4. Find the density function \(f_U(u)\) by differentiating \(F_U(u)\). Thus, \(f_U(u) = \dfrac{dF_U(u)}{du}\).

Example: The Method of Distribution Functions

Example: Let \(Y\) have probability density function given by \[ f_Y(y) = \begin{cases} \frac{y+1}{2}, & -1 \le y \le 1, \\ 0, & \text{otherwise.} \end{cases} \]

The Method of Transformations

We can take the ideas in that last example further.

Summary of the Transformation Method

Let \(U = h(Y)\), where \(h(y)\) is either an increasing or decreasing function of \(y\) for all \(y\) such that \(f_Y(y) > 0\).

1. Find the inverse function, \(y = h^{-1}(u)\).
2. Evaluate \(\dfrac{dh^{-1}}{du} = \dfrac{d[h^{-1}(u)]}{du}\).
3. Find \(f_U(u)\) by \[ f_U(u) = f_Y[h^{-1}(u)]\left\lvert \frac{dh^{-1}}{du} \right\rvert. \]

Example: The Method of Transformations

Example: Let \(Y\) have the probability density function given by: \[ f(y) = \begin{cases} 2y, & 0 \le y \le 1, \\ 0, & \text{otherwise.} \end{cases} \] Find the density function of \(U = -4Y+3\).

Example: The Method of Transformations

Example: Consider a random variable \(Y_1\), the proportion of impurities in a chemical sample, and \(Y_2\), the proportion of type I impurities among all impurities in the sample. The joint density function was given by \[ f(y_1, y_2) = \begin{cases} 2(1 - y_1), & 0 \le y_1 \le 1, 0 \le y_2 \le 1, \\ 0, & \text{otherwise.} \end{cases} \] We are interested in \(U = Y_1 Y_2,\) which is the proportion of type I impurities in the sample. Find the probability density function for \(U\).

Multivariable Transformations Using Jacobians

The Bivariate Transformation Method

Suppose that \(Y_1\) and \(Y_2\) are continuous random variables with joint density function \(f_{Y1,Y2}(y_1, y_2)\) and that for all \((y_1, y_2)\), such that \(f_{Y1,Y2}(y_1, y_2) > 0\), \[\begin{align} u_1 &= h_1(y_1, y_2), & &\text{and} & u_2 &= h_2(y_1, y_2), \end{align}\] is a one-to-one transformation from \((y_1, y_2)\) to \((u_1, u_2)\) with inverse \[\begin{align} y_1 &= h^{-1}_1(u_1, u_2), & &\text{and} & y_2 &= h^{-1}_2(u_1, u_2). \end{align}\]

Multivariable Transformations Using Jacobians

The Bivariate Transformation Method

If \(h^{−1}_1 (u_1, u_2)\) and \(h^{−1}_2 (u_1, u_2)\) have continuous partial derivatives with respect to \(u_1\) and \(u_2\) and Jacobian \[ J = \text{det} \begin{bmatrix} \frac{\partial h^{-1}_1}{\partial u_1} & \frac{\partial h^{-1}_1}{\partial u_2} \\ \frac{\partial h^{-1}_2}{\partial u_1} & \frac{\partial h^{-1}_2}{\partial u_2} \\ \end{bmatrix} = \left( \frac{\partial h^{-1}_1}{\partial u_1} \right) \left( \frac{\partial h^{-1}_2}{\partial u_2} \right) - \left( \frac{\partial h^{-1}_2}{\partial u_1} \right) \left( \frac{\partial h^{-1}_1}{\partial u_2} \right) \ne 0, \] Then the joint density of \(U_1\) and \(U_2\) is \[ f_{U_1, U_2}(u_1, u_2) = f_{Y_1, Y_2}\Big( h^{-1}_1(u_1, u_2), h^{-1}_2(u_1, u_2) \Big) \lvert J \rvert, \] where \(\lvert J \rvert\) is the absolute value of the Jacobian.

Example: Multivariable Transformations Using Jacobians

Example: Consider random variables \(Y_1\) and \(Y_2\), with joint density function \[ f(y_1, y_2) = \begin{cases} 2(1 - y_1), & 0 \le y_1 \le 1, 0 \le y_2 \le 1, \\ 0, & \text{otherwise.} \end{cases} \] If \(U = Y_1 Y_2\), find the probability density function for \(U\) by using the bivariate transformation method.