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Apply Transformations

Apply Transformations

Transformations allow us to find the distribution of a function of random variables. There are different methods of applying transformations.

The Method of Distribution Function

Given a random variable \(Y\) that is a function of a random variable \(X\), that is  \(Y=u(X)\), we can find the cumulative density function of \(Y\), that is \(G(y)\), directly and then differentiate this cumulative distribution function (CDF) to get the PDF of \(Y\), that is \(g(y)\):

$$ \begin{align*} G\left(y\right)&=P\left(Y\le y\right)=P[u\left(X\right)\le y] \\ g\left(y\right)&=G^\prime\left(y\right) \end{align*} $$

Example: The Method of Distribution Function

A random variable \(X\) has a uniform distribution over the interval \([-2,2]\), and another variable \(Y\) is defined as \(Y=X^2\).

Find the probability density function of \(Y\), \(f(y)\).

Solution

We first find the cumulative density function of \(Y\), that is \(G(y)\):

$$\begin{align}G(y)&=P(Y\leq y)\\ & =P(X^2\leq y)\\ &=P(X\leq \sqrt{y})\\ &=P(-\sqrt {y}\le X\le \sqrt {y}\\ &=\int_{-\sqrt y}^{\sqrt y}{f\left(x\right)dx=2*\int_{0}^{\sqrt y}{\frac{1}{4}dx=2*\frac{1}{4}\left[x\right]_0^{\sqrt y}}}\\ &=\frac{\sqrt y}{2}\end{align}$$

Thus,

$$g\left(y\right)=\frac{d}{dy}\left(\sqrt y\right)=\frac{1}{4\sqrt y}\ ,\ for-2\le y\le 2$$

The Method of Transformation 

In this method, we consider a variable \(X\), whose PDF, \(f(x)\), is given and another variable \(Y\), such that, \(Y=u(X) \) where \(u(x)\) is either an increasing or decreasing function of \(x\) that is \(u\left(x\right) > 0\).

We first find the inverse function of  \(u\left(x\right)\), that is \(x=u^{-1}\left(y\right)\).

We then evaluate:

$$\frac{du^{-1}}{dy}=\frac{d\left[u^{-1}\left(y\right)\right]}{dy}$$

We then finally find \(f(y)\), using the formula:

$$f_Y\left(y\right)=f_X\left[u^{-1}\left(y\right)\right]\left|\frac{du^{-1}}{dy}\right|$$

Example: Applying Transformation – Use of the Inverse Function

Let \(X\) be a random variable with a PDF:

$$f(x)=\begin{cases} x^2 +\frac{2}{3},&0 < x < 1\\ 0, & \text{otherwise}\end{cases}$$

Find the PDF of \(U=2x-3\).

Solution

Let \(z(X)=2x-3\).

If \(u=2x-3\); and making \(x\) the subject of the formula:

$$x=z^{-1}(u)=\frac{u+3}{2}$$

And thus,

$$\frac{{dz}^{-1}(u)}{du}=\frac{dx}{du}=\frac{1}{2}$$

Now,

$$\begin{align}f\left(u\right)&=f_X\left(z^{-1}\left(u\right)\right)\left|\frac{dx}{du}\right|\\&=\begin{cases} [(u+3/2)^2+2/3]|1/2|=1/2 (u+3/2)^2+1/3, &-3 < u < 1\\0, & \text{elsewhere}\end{cases} \end{align}$$

Change of Variable Technique

We can use the change-of-variable technique to  find the PDF of a transformed variable, \(Y\).

$$ g\left( y \right) =f\left[ \upsilon \left( y \right) \right] |{ \upsilon }^{ \prime }\left( y \right) | $$

where \(\upsilon \left( y \right)\) is the inverse function of \(y\).

Example: Change-of-variable Technique

Given the following probability density function of a continuous random variable:

$$ f\left( x \right) =\begin{cases} { x }^{ 2 }+\cfrac { 2 }{ 3, } & 0< x < 1 \\ 0, & \text{otherwise} \end{cases} $$

Let \(Y = X – 50\).

Find \(g(y)\).

Solution

\(v\left(y \right) = Y + 50\)

\(v^{\prime} \left(y \right) = 1\)

$$ g\left( y \right) =f\left[ \upsilon \left( y \right) \right] \left[ { \upsilon }^{ \prime }\left( y \right) \right] =\left[ { \left( Y+50 \right) }^{ 2 }+\frac { 2 }{ 3 } \right] \left[ 1 \right] ={ \left( Y+50 \right) }^{ 2 }+\frac { 2 }{ 3 } $$

The PDF of a discrete transformed random variable can be found in a similar manner, as shown in the formula below:

$$ g\left( y \right) =f\left[ \upsilon \left( y \right) \right] $$

Note: \(|{ \upsilon }^{ \prime }\left( y \right) |\) is not needed in this case.

Learning Outcome:

Topic 2g: Univariate Random Variables – Apply transformations.

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