# Functions of Random Variables

Tags
Statistics
Visualization
Date
Dec 8, 2019
Description
Finding the distribution of a real-valued function of multiple random variables. There's the method of distribution functions, transformations and moment generating functions.
Slug
mathematical-statistics-functions-of-random-variables
Suppose we have random variables , and a real-valued function . In this chapter, weβre only going to do one thing: introduce multiple methods to find the probability distribution of . Any one of these methods can be employed to find the distribution , but usually one of the methods leads to a simpler derivation than the others. The βbestβ method varies from one application to another.

### The method of distribution functions

The first way of finding the distribution of is using the definition directly.
We can work this out in the following steps:
1. Write out the distribution function of :
1. Find the region of such that and denote the region as .
1. Integrate over .
The hardest part of this method is finding the set . Weβll gain some insight with some examples.

#### Sugar example

β
A company is selling sugar online. Suppose the amount of sugar it can sell per day is tons, which is a continuous random variable with density function defined as
For each ton of sugar sold, the company can earn $300. The daily operation cost is$100. Find the probability distribution of the daily income of this company.
Let the random variable denote the daily profit in hundred dollars. We want to write as and find .
If , . If , . When
Note that as ranges from 0 to 1, ranges from β-1 to 2, so the distribution function of is
The density function can also be calculated:

#### Example of two variables

Suppose and are two continuous random variables with joint density function
Find the density function of . Also use the density function of to find .
We first need to find and use it to obtain the density function .
Now we need to find the region of and . We know that and , and is bounded between 0 and 1 due to the latter condition.
R code.
library(ggplot2)

ggplot(NULL, aes(x = c(0, 1))) +
stat_function(
fun = ~ .x - 0.3, geom = "area", xlim = c(0.3, 1), fill = "#0073C2", alpha = 0.7
) +
stat_function(
fun = ~ .x, geom = "area", xlim = c(0, 1), fill = "#CD534C", alpha = 0.5
) +
geom_segment(aes(x = 1, y = 0, xend = 1, yend = 1), linetype = "dashed") +
geom_segment(aes(x = 0, y = 1, xend = 1, yend = 1), linetype = "dashed") +
labs(x = expression(y[1]), y = expression(y[2])) +
ggpubr::theme_pubr()
We can find the integral by subtracting the lower-right blue triangle region () from the entire red triangle.
So the distribution function of is
Calculating the density function of is now straightforward:
which gives us
The expectation of can be calculated as

### Sum of independent random variables

An important application of the method of distribution functions is to calculate the distribution of from the distributions of and when they are independent, continuous random variables.
The first few steps are the same as whatβs described above:
Define the region such that
We have
Now the independence comes into place:
The distribution function of is called the convolution of and . By differentiating the above distribution function, we can find the density function of .

#### Uniform distribution example

If and are two independent random variables both uniformly distributed on , calculate the probability density of . We can directly apply the equation above:
We know that and . There are several cases here:
1. If , , then .
1. If , we also need , or .
1. If , we also need , or .
1. If , .
When
When
In summary,
The sum of two uniform random variables is called a triangular random variable because of the shape of the density function above.
R code.
ggplot(NULL, aes(x = c(-1, 3))) +
stat_function(fun = ~ 0, geom = "line", xlim = c(-1, 0)) +
stat_function(fun = ~ .x, geom = "line", xlim = c(0, 1)) +
stat_function(fun = ~ 2 - .x, geom = "line", xlim = c(1, 2)) +
stat_function(fun = ~ 0, geom = "line", xlim = c(2, 3)) +
labs(x = "a", y = expression(f[X+Y](a))) +
ggpubr::theme_pubr()

#### Normal distribution example

Suppose and are two independent standard normal random variables. Find the density of . Recall that
Therefore is a normal random variable with mean 0 and variance 2. Similar results of this example can be obtained for a more general case, which is a very important property of normal random variables.
Theorem
Let be a sequence of independent normal random variables with parameters and , then is normally distributed with parameters and .

### The method of transformations

We first consider the univariate case. Suppose is a continuous random variable with density function . To find
what we can do is to transform the condition back to a condition of . We know that exists when the function is monotonic, i.e.Β , .
Given is a monotonic function of , maps every distinct value of to a distinct value of .
If is a monotonically increasing function,
If is a monotonically decreasing function,
These two cases can be unified as follows using the difference in the sign of . If is a monotonic function for all such that ,

#### Sugar example revisited

β
In the sugar example, we defined the density function of as
and . Find with the method of transformations.
When
so is a monotonically decreasing function. The inverse function is
and the first derivative is
Now we can find the density function using the formula
Clean this up a bit and we get

#### Multivariate example

The transformation method can also be used in multivariate situations. Let random variables and have a joint density function
Find the density function for .
We can prove that and use the method described earlier to find the distribution for the sum of two independent random variables. We can also apply the method of transformations here. If we fix , we have
From here we can consider this as a one-dimensional transformation problem.
Which is
Using the joint density function of and , we can obtain the marginal density function of :

#### Multivariate procedure

As shown in the example above, when our problem is and we want to find , the procedure is
1. Fix , and denote
1. Calculate the joint density function of and using the formula
1. Find the marginal density of with
β
Suppose random variables and have a joint density function
Find the density function of .
Following the procedure, we first fix for some . Then we consider the univariate transformation and get the joint density function for and .
The joint density function of and is
and now we can obtain the marginal density of :

### The method of moment generating functions

We know that for a random variable , its associated moment generating functions is given by . The moment can help us find .
For random variables and , if both moment generating functions exist and , we have , i.e.Β  and have the same probability distribution.

#### Normal random variables

Let and . Show that using the method of moment generating functions.
Define . If we can show that , then the distribution of is identical to the distribution of .
We know that , so

#### Use for sum of independent random variables

The method of moment generating functions is also very useful for calculating the probability distributions for the sum of independent random variables.
Suppose and are independent with moment generating functions and . If , then
In this special case,

#### Some more examples

β
Find the distribution of if and are
1. independent binomial random variables with parameters and , respectively.
1. independent Poisson random variables with parameters and , respectively.
In the first scenario we have and . since . Recall that
so the MGF of is
which tells us .
In the second case, we have and . Since
the MGF of can be written as
So .