Functions of a R.V. - Transformations

Example

say X~U[0,1] and λ>0. What is the pdf of the random variable Y=-λlog(X)?

Solution: we first find the cdf and then the pdf as follows:

if y>0. For y<0 note that P(-logX<y) = 0 because 0<X<1, so logX<0, so -logX>0 always.

This is an example of a function (or transformation) of a random variable. These transformations play a major role in probability and statistics. We will see how to find their pdf’s on a few example**.

Example

Say X is the number of roles of a fair die until the first six. We have already seen that P(X=x) = 1/6*(5/6)^x-1, x=1,2,.. Let Y be 1 if X is even, 0 otherwise. Find the of Y
Note: here both X and Y are discrete.
Solution:

and P(Y=0) = 1 - P(Y=1) = 5/11

Example

say X is a continuous r.v with pdf f_X(x) = 1/2exp(-|x|) \(x\in \mathbb{R}\) (this is called a double exponential) Let Y=I_[-1,1](X). Find the of Y.

Note: here X is continuous and Y is discrete.

Example

again let X have pdf f_X(x) = 1/2exp(-|x|) \(x\in \mathbb{R}\). Let Y =X². Then for y<0 we have P(Y≤y) = 0. So let y>0. Then

Next up some examples of functions of random vectors:

Example

say (X,Y) is a bivariate standard normal r.v, that is it has joint density given by

for \((x,y) \in \mathbb{R}^2\).

Let the r.v. (U,V) be defined by U = X+Y and V = X-Y. Find the joint pdf of (U,V).

To start let’s define the functions g₁(x,y) = x+y and g₂(x,y) = x-y, so that U=g₁(X,Y) and V = g₂(X,Y).

For what values of u and v is f_(U,V)(u,v) positive? Well, for any values for which the system of 2 linear equations in two unknowns u=x+y and u=x-y has a solution. These solutions are

x = h₁(u,v) = (u + v)/2
y = h₂(u,v) = (u - v)/2

From this we find that for any \((u,v) \in \mathbb{R}^2\) there is a unique \((x,y) \in \mathbb{R}^2\) such that u=x+y and v=x-y. So the transformation \((x,y) \rightarrow (u,v)\) is one-to-one and therefore has a Jacobian given by

Now from multivariable calculus we have the following:

Note that the density factors into a function of u and a function of v.

This is not only a necessary but also a sufficient condition for U and V to be independent.

Example

say X and Y are independent standard normal r.v.’s. Let Z = X + Y. Find the pdf of Z.

Note: now we have a transformation from \(\mathbb{R}^2 \rightarrow \mathbb{R}\).

Note: Z = X + Y = U in the example above, so the pdf of Z is just the marginal of U and we find

Say X and Y are two continuous independent r.v with pdf f_X and f_Y, and let Z = X+Y. If we repeat the above calculations we can show that in general the pdf of Z is given by

This is called the convolution formula.

There is a second method for deriving the convolution formula which is useful. It uses a continuous analog to the law of total probability:

In the setup from above we have

The tricky part of this is the interchange of the derivative and the integral. Working with densities and cdfs usually means they are ok.

Example

Say X₁, .., X_n are iid U[0,1]. Let M=max{X₁, .., X_n}. Find f_M.

Note

so we see that M~Beta(n,1), from which is follows for example that EM=n/(n+1)

M is an example of an order statistic. In general the k^th largest observation of a sample is called the k^th order statistic