Transformations
Dear students,
Since we have covered the mgf technique extensively already, here we only review the cdf and the pdf techniques, first for univariate (one-to-one and more-to-one) and then for bivariate (one-to-one and more-to-one) transformations.
1. The cumulative distribution function (cdf) technique
Suppose Y is a continuous random variable with cumulative distribution function (cdf) . Let be a function of Y, and our goal is to find the distribution of U. The cdf technique is especially convenient when the cdfhas closed form analytical expression. This method can be used for both univariate and bivariate transformations.
Steps of the cdf technique:
- Identify the domain of Y and U.
- Write, the cdf of U, in terms of , the cdf of Y .
- Differentiate to obtain the pdf of U,
Example 1.Suppose that Find the distribution of .
Solution.The cdf of is given by
The domain (*domain is the region where the pdf is non-zero) for is; thus, because , it follows that the domain for U is .
The cdf of U is:
Because for 0 < y < 1; i.e., for u > 0, we have
Taking derivatives, we get, for u > 0,
Summarizing,
This is an exponential pdf with mean; that is, U ~ exponential(λ = 1). □
Example 2.Suppose that. Find the distribution of the random variable defined by U = g(Y ) = tan(Y ).
Solution. The cdf of is given by
The domain for Y is . Sketching a graph of the tangent function from to , we see that .
Thus, , the set of all reals. The cdf of U is:
Because , we have
The pdf of U, for , is given by
Summarizing,
A random variable with this pdf is said to have a (standard) Cauchy distribution. One interesting fact about a Cauchy random variable is that none of its moments are finite. Thus, if U has a Cauchy distribution, E(U), and all higher order moments, do not exist.
Exercise: If U is standard Cauchy, show that. □
2. The probability density function (pdf) technique, univariate
Suppose that Y is a continuous random variable with cdf and domain, and let , where is a continuous, one-to-onefunction defined over . Examples of such functions include continuous (strictly) increasing/decreasing functions. Recall from calculus that if is one-to-one, it has an unique inverse . Also recall that if is increasing (decreasing), then so is .
Derivation of the pdf technique formula using the cdf method:
Suppose that is a strictly increasing functionof y defined over . Then, it follows that and
Differentiating with respect to u, we get
Now as is increasing, so is ; thus, If is strictly decreasing, then
and , which gives
Combining both cases, we have shown that the pdf of U, where nonzero, is given by
It is again important to keep track of the domain for U. If denotes the domain of Y, then , the domain for U, is given by
Steps of the pdf technique:
1. Verify that the transformation u = g(y) is continuous and one-to-one over .
2. Find the domains of Y and U.
3. Find the inverse transformation and its derivative (with respect to u).
4. Use the formula above for .
Example 3.Suppose that Y ~ exponential(β); i.e., the pdf of Y is
Let , . Use the method of transformations to find the pdf of U.
Solution.First, we note that the transformation is a continuous strictly increasing function of y over , and, thus, is one-to-one. Next, we need to find the domain of U. This is easy since y > 0 implies as well.
Thus, . Now, we find the inverse transformation:
and its derivative:
Thus, for u > 0,
Summarizing,
This is a Weibull distribution. The Weibull family of distributions is common in life science (survival analysis), engineering and actuarialscience applications. □
Example 4. Suppose that Y ~ beta(α= 6; β= 2); i.e., the pdf of Y is given by
What is the distribution of U = g(Y ) = 1Y ?
Solution. First, we note that the transformation g(y) = 1Y is a continuous decreasing function of y over , and, thus, g(y) is one-to-one. Next, we need
to find the domain of U. This is easy since 0 < y < 1 clearly implies 0 < u < 1. Thus,
. Now, we find the inverse transformation:
and its derivative:
Thus, for 0 < u < 1,
Summarizing,
We recognize this is a beta distribution with parameters α= 2 and β= 6.□
More-to-one transformation: What happens if u = g(y) is not a one-to-one transformation? In this case, we can still use the method of transformations, but we have “break up" the transformation into disjoint regions where g is one-to-one.
RESULT: Suppose that Y is a continuous random variable with pdf and that U =g(Y ), not necessarily a one-to-one (but continuous) function of y over RY. However,suppose that we can partition into a finite collection of sets, say, A0,A1, A2,…,Ak, wherefor all i ≠ 0, and is continuous on each Ai ,i ≠ 0. Furthermore, suppose thatthe transformation is 1-to-1 from Ai(i= 1, 2,…,k,) to B, where B is the domain of U = g(Y )such that is a 1-to-1 inverse mapping of Y to U = g(Y )from B toAi.
Then, the pdf of U is given by
That is, writing the pdf of U can be done by adding up the terms corresponding to each disjoint set Ai, i= 1, 2,…,k.
Example 5.Suppose that Y ~ N(0, 1); that is, Y has a standard normal distribution; i.e.,
Consider the transformation: .
Solution 1 (the pdf technique):
This transformation is not one-to-one on , but it is one-to-one on A1 = (0) and A2 = (0,) (separately) since is decreasing on A1 and increasing on A2, and A0= {0} where . Furthermore, note that A0, A1 and A2 partitions . Summarizing,
Partition / Transformation / Inverse transformationA1 = (0) / /
A2 = (0,) / /
And, on both sets A1 and A2,
Clearly, ; thus, , and the pdf of U is given by
Thus, for u > 0, and recalling that collapses to
Summarizing, the pdf of U is
That is, U ~ gamma(1/2, 2). Recall that the gamma(1/2, 2) distribution is the same as a distribution with 1 degree of freedom; that is, . □
Solution 2 (the cdf technique):
Taking derivative with respect to at both sides, we have:
That is, U ~ gamma(1/2, 2). Recall that the gamma(1/2, 2) distribution is the same as a distribution with 1 degree of freedom; that is, . □
3. The probability density function (pdf) technique, bivariate
Here we discuss transformations involving two random variable . Thebivariate transformation is
Assuming that are jointly continuousrandom variables, we will discuss the one-to-one transformation first. Starting with the joint distribution of , our goal is to derive the joint distribution of .
Suppose that is a continuous random vector with joint pdf. Let be a continuous one-to-one vector-valued mapping from to , where and , and where and denote the two-dimensional domain of and , respectively. If and have continuous partial derivatives with respect to both , and the Jacobian, J, where, with “det” denoting “determinant”,
then
where|J|denotes the absolute value of J.
RECALL: The determinant of a 2 2 matrix, e.g.,
Steps of the pdf technique:
1. Find , the joint distribution of . This may be given in the problem. If Y1 and Y2 are independent, then .
2. Find , the domainof .
3. Find the inverse transformations and .
4. Find the Jacobian, J, of the inverse transformation.
5. Use the formula above to find , the joint distribution of .
NOTE: If desired, marginal distributions . can be found by integratingthe joint distribution .
Example 6.Suppose that gamma(α, 1), gamma(β, 1), and thatare independent. Define the transformation
Find each of the following distributions:
(a) , the joint distribution of ,
(b) , the marginal distribution of , and
(c), the marginal distribution of .
Solutions.(a) Since are independent, the joint distribution of is
forand 0, otherwise. Here, By inspection, we see that , and must fall between 0 and 1.
Thus, the domain of is given by
The next step is to derive the inverse transformation. It follows that
The Jacobian is given by
We now write the joint distribution for . For , we have that
Note: We see that are independent since the domain does not constrain by or vice versa and since the nonzero part
of can be factored into the two expressions and , where
and
(b) To obtain the marginal distribution of , we integrate the joint pdf
over . That is, for 0,
Summarizing,
We recognize this as a gamma(1) pdf; thus, marginally, ~gamma(1).
(c) To obtain the marginal distribution of, we integrate the joint pdf over . That is, for ,
Summarizing,
Thus, marginally, U2 ~ beta().□
REMARK: Suppose that is a continuous random vector with joint pdf , and suppose that we would like to find the distribution of a single random variable
Even though there is no present here, the bivariate transformation technique can still be useful. In this case, we can devise an “extra variable” , perform the bivariate transformation to obtain , and then find the marginal distribution of by integrating out over the dummy variable . While the choice of is arbitrary, there are certainly bad choices. Stick with something easy; usually does the trick.
Exercise: (Homework 3, Question 1)Suppose that and are random variables with joint pdf
Find the pdf of .
More-to-one transformation: What happens if the transformation of Y to U is not a one-to-one transformation? In this case, similar to the univariate transformation, we can still use the pdf technique, but we have to “break up" the transformation into disjoint regions where gis one-to-one.
RESULT: Suppose that Yis a continuous bivariate random variable with pdf and that , , where is not necessarily a one-to-one (but continuous) function of yover RY= . Furthermore, suppose that we can partition into a finite collection of sets, say, A0, A1, A2,…,Ak, where for all i ≠ 0, and is continuous on each Ai ,i ≠ 0. Furthermore, suppose that the transformation is 1-to-1 from Ai(i= 1, 2,…,k,) to B, where B is the domain of such that is a 1-to-1 inverse mapping of Y to Ufrom B toAi.
Let Ji denotes the Jacobina computed from the ith inverse, i= 1, 2,…,k. Then, the pdf of Uis given by
Example 7.Suppose that N(0, 1), N(0, 1), and thatare independent. Define the transformation
Find each of the following distributions:
(a) , the joint distribution of ,
(b) , the marginal distribution of
Solutions.(a) Since are independent, the joint distribution of is
Here,
The transformation of Y to U is not one-to-one because the points and are both mapped to the same point. But if we restrict considerations to either positive or negative values of , then the transformation is one-to-one. We note that the three sets below form a partition of as defined above with , , and .
The domain of U, is the image of both and under the transformation. The inverse transformation from and are given by:
and
The Jacobians from the two inverses are
The pdf of U on its domain B is thus:
Plugging in, we have:
Simplifying, we have:
(b) To obtain the marginal distribution of , we integrate the joint pdf
over . That is,
Thus, marginally, U1follows the standardCauchy distribution.□
REMARK: The transformation method can also be extended to handle n-variate transformations. Suppose that are continuous random variables with joint pdf
and define
Example 8. Given independent random variables and, each with uniform distributions on (0, 1), find the joint pdf of U and Vdefined by U=X+Y, V=X-Y, and the marginal pdf of U.
The joint pdf of and is.
The inverse transformation, written in terms of observed values is
It is clearly one-to-one. The Jacobian is
, so
We will use to denote the range space of and to denote that of and these are shown in the diagrams below. Firstly, note that there are 4 inequalities specifying ranges ofand, and these give 4 inequalities concerning and, from which can be determined. That is,
that is,
that is
, that is
, that is
Drawing the four lines
On the graph, enables us to see the region specified by the 4 inequalities.
Now, we have
The importance of having the range space correct is seen when we find marginal pdf of
, using indicator functions.
Example 9. Given and are independent random variables each with pdf, find the distribution of.
We note that the joint pdf of and is
Define. Now we need to introduce a second random variable which is a function of and. We wish to do this in such a way that the resulting bivariate transformation is one-to-one and our actual task of finding the pdf of U is as easy as possible. Our choice for is of course, not unique. Let us define. Then the transformation is, (using, since we are really dealing with the range spaces here).
From it, we find the Jacobian,
To determine, the range space of and, we note that
, that is
So is as indicated in the diagram below.
Now, we have
The marginal pdf of is obtained by integrating with respect to, giving
[This is sometimes called the distribution.]
Homework #3.
Question 1 is the exercise on page 9 of this handout.
Questions 2-9 (from our textbook): 2.9, 2.15, 2.24, 2.30, 2.31, 2.32, 2.33, 2.34
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