18  Distributions of Transformations of Random Variables

• A function of a random variable is a random variable: if \(X\) is a random variable and \(g\) is a function then \(Y=g(X)\) is a random variable.
• Since \(g(X)\) is a random variable it has a distribution. In general, the distribution of \(g(X)\) will have a different shape than the distribution of \(X\).

18.1 Linear rescaling

Example 18.1

Let \(U\) have a Uniform(0, 1) distribution. Define \(V=1-U\).

1. Does \(V\) result from a linear rescaling of \(U\)?
   
   
   
2. What are the possible values of \(V\)?
   
   
   
3. Is \(V\) the same random variable as \(U\)?
   
   
   
4. Find \(\text{P}(U \le 0.1)\) and \(\text{P}(V \le 0.1)\).
   
   
   
5. Sketch a plot of what the histogram of many simulated values of \(V\) would look like.
   
   
   
6. Does \(V\) have the same distribution as \(U\)?
   
   
   

Example 18.2

Suppose that \(X\), the SAT Math score of a randomly selected student, follows a Normal(500, 100) distribution. Randomly select a student and let \(X\) be the student’s SAT Math score. Now have the selected student spin the Normal(0, 1) spinner. Let \(Z\) be the result of the spin and let \(Y=500 + 100 Z\).

1. Is \(Y\) the same random variable as \(X\)?
   
   
   
2. Does \(Y\) have the same distribution as \(X\)?
   
   
   

Example 18.3

Let \(Z\) be a random variable with a Normal(0, 1) distribution. Consider \(-Z\).

1. Donny Don’t says that the distribution of \(-Z\) will look like an “upside-down bell”. Is Donny correct? If not, explain why not and describe the distribution of \(-Z\).
   
   
   
2. Donny Don’t says that the standard deviation of \(-Z\) is -1. Is Donny correct? If not, explain why not and determine the standard deviation of \(-Z\).
   
   
   

• A linear rescaling is a transformation of the form \(g(u) = a + bu\).
• A linear rescaling of a random variable does not change the basic shape of its distribution, just the range of possible values.
  • However, remember that the possible values are part of the distribution. So a linear rescaling does technically change the distribution, even if the basic shape is the same. (For example, Normal(500, 100) and Normal(0, 1) are two different distributions.)
• A linear rescaling transforms the mean in the same way the individual values are transformed.
• Adding a constant to a random variable does not affect its standard deviation.
• Multiplying a random variable by a constant multiplies its standard deviation by the absolute value of the constant.
• Whether in the short run or the long run, \[\begin{align*} \text{Average of $a+bX$} & = a+b(\text{Average of $X$})\\ \text{SD of $a+bX$} & = |b|(\text{SD of $X$})\\ \text{Variance of $a+bX$} & = b^2(\text{Variance of $X$}) \end{align*}\]
• If \(U\) has a Uniform(0, 1) distribution then \(X = a + (b-a)U\) has a Uniform(\(a\), \(b\)) distribution.
• If \(Z\) has a Normal(0, 1) distribution then \(X = \mu + \sigma Z\) has a Normal(\(\mu\), \(\sigma\)) distribution.
• Remember, do NOT confuse a random variable with its distribution.
  • The random variable is the numerical quantity being measured
  • The distribution is the long run pattern of variation of many observed values of the random variable

18.2 Nonlinear transformations of random variables

• In general the shape of the distribution \(g(X)\) will be different than that of \(X\).
• In general: whether in the short run or the long run

\[ \text{Average of } g(X) \neq g(\text{Average of }X). \]

Example 18.4

Let \(U\) have a Uniform(0, 1) distribution, and let \(X = -\log(1-U)\). For each of the intervals in the table below find the probability that \(U\) lies in the interval, and identify the corresponding values of \(X\). (You should at least compute a few by hand to see what’s happening, but you can use software to fill in the rest.) After completing the table, sketch a histogram representing the distribution of \(X\).

Interval of U	Length of U interval	Probability	Interval of X	Length of X interval
(0, 0.1)
(0.1, 0.2)
(0.2, 0.3)
(0.3, 0.4)
(0.4, 0.5)
(0.5, 0.6)
(0.6, 0.7)
(0.7, 0.8)
(0.8, 0.9)
(0.9, 1)

Example 18.5

For each of the intervals of \(X\) values in the table below identify the corresponding values of \(U\), and then find the probability that \(X\) lies in the interval. (You should at least compute a few by hand to see what’s happening, but you can use software to fill in the rest.) After completing the table, sketch a histogram representing the distribution of \(X\). Hint: if \(X = -\log(1-U)\) then \(U = 1-e^{-X}\).

Interval of X	Length of X interval	Probability	Interval of U	Length of U interval
(0, 0.5)
(0.5, 1)
(1, 1.5)
(1.5, 2)
(2, 2.5)
(2.5, 3)
(3, 3.5)
(3.5, 4)
(4, 4.5)
(4.5, 5)

Example 18.6

We have now seen a few reasons why if \(U\) has a Uniform(0, 1) distribution then \(X=-\log(1-U)\) has an Exponential(1) distribution. The purpose of the example is derive the pdf of \(X\), starting from just the setup in the previous sentence.

1. Identify the possible values of \(X\). (We have done this already, but this should always be your first step.)
   
   
   
   
   
2. Let \(F_X\) denote the cdf of \(X\). Find \(F_X(1)\).
   
   
   
   
   
3. Find \(F_X(2)\).
   
   
   
   
   
4. Find the cdf \(F_X(x)\).
   
   
   
   
   
5. Find the pdf \(f_X(x)\).
   
   
   
   
   
6. Why should we not be surprised that \(X=-\log(1-U)\) has cdf \(F_X(x) = 1 - e^{-x}\)? Hint: what is the function \(u\mapsto -\log(1-u)\) in this case?
   
   
   
   
   

• If \(X\) is a continuous random variable whose distribution is known, the cdf method can be used to find the pdf of \(Y=g(X)\)
• Determine the possible values of \(Y\). Let \(y\) represent a generic possible value of \(Y\).
• The cdf of \(Y\) is \(F_Y(y) = \text{P}(Y\le y) = \text{P}(g(X) \le y)\).
• Rearrange \(\{g(X) \le y\}\) to get an event involving \(X\). Warning: it is not always \(\{X \le g^{-1}(y)\}\). Sketching a picture of the function \(g\) helps.
• Obtain an expression for the cdf of \(Y\) which involves \(F_X\) and some transformation of the value \(y\).
• Differentiate the expression for \(F_Y(y)\) with respect to \(y\), and use what is known about \(F'_X = f_X\), to obtain the pdf of \(Y\). You will typically need to apply the chain rule when differentiating.
• You will need to use information about \(X\) at some point in the last step above. You can either:
  • Plug in the cdf of \(X\) and then differentiate with respect to \(y\).
  • Differentiate with respect to \(y\) and then plug in the pdf of \(X\).
  • Either way gets you to the correct answer, but depending on the problem one way might be easier than the other.

Example 18.7

Let \(X\) be a random variable with a Uniform(-1, 1) distribution and Let \(Y=X^2\).

1. Identify the possible values of \(Y\).
   
   
   
   
   
2. Sketch the pdf of \(Y\). Hint: consider a few equally spaced intervals of \(Y\) values and see what \(X\) values they correspond to.
   
   
   
   
   
3. Run a simulation to approximate the pdf of \(Y\).
   
   
   
   
   
4. Find \(F_Y(0.49)\).
   
   
   
   
   
5. Use the cdf method to find the pdf of \(Y\). Is the pdf consistent with your simulation results?