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# 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)&amp;=P(Y\leq y)\\ &amp; =P(X^2\leq y)\\ &amp;=P(X\leq \sqrt{y})\\ &amp;=P(-\sqrt {y}\le X\le \sqrt {y}\\ &amp;=\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}}}\\ &amp;=\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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