# Linear Transformation of Continuous Random Variable is Continuous Random Variable

## Theorem

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.

Let $a$ be a non-zero real number.

Let $b$ be a real number.

Let $X$ be a continuous real variable.

Let $F_X$ be the cumulative distribution function of $X$.

Then $a X + b$ is a continuous real variable.

Further, if $a > 0$, the cumulative distribution function of $a X + b$, $F_{a X + b}$. is given by:

$\ds \map {F_{a X + b} } x = \map {F_X} {\frac {x - b} a}$

for each $x \in \R$.

If $a < 0$, $F_{a X + b}$ is given by:

$\ds \map {F_{a X + b} } x = 1 - \map {F_X} {\frac {x - b} a}$

for each $x \in \R$.

## Proof

Since $X$ is a continuous real variable, we have that:

$F_X$ is continuous.

We use this fact to show that $F_{a X + b}$ is continuous, showing that $a X + b$ is a continuous real variable.

We split up into cases.

Suppose that $a > 0$.

Then, for each $x \in \R$, we have:

 $\ds \map {F_{a X + b} } x$ $=$ $\ds \map \Pr {a X + b \le x}$ $\ds$ $=$ $\ds \map \Pr {a X \le x - b}$ $\ds$ $=$ $\ds \map \Pr {X \le \frac {x - b} a}$ $\ds$ $=$ $\ds \map {F_X} {\frac {x - b} a}$ Definition of Cumulative Distribution Function

From Composite of Continuous Mappings is Continuous and Linear Function is Continuous, we therefore have:

$F_{a X + b}$ is continuous

in the case $a > 0$.

Now suppose that $a < 0$.

Then, for each $x \in \R$, we have:

 $\ds \map {F_{a X + b} } x$ $=$ $\ds \map \Pr {a X + b \le x}$ $\ds$ $=$ $\ds \map \Pr {a X \le x - b}$ $\ds$ $=$ $\ds \map \Pr {X \ge \frac {x - b} a}$ $\ds$ $=$ $\ds \map \Pr {\Omega \setminus \set {X < \frac {x - b} a} }$ $\ds$ $=$ $\ds 1 - \map \Pr {X < \frac {x - b} a}$ Probability of Event not Occurring $\ds$ $=$ $\ds 1 - \map \Pr {X \le \frac {x - b} a}$ Probability of Continuous Random Variable Lying in Singleton Set is Zero $\ds$ $=$ $\ds 1 - \map {F_X} {\frac {x - b} a}$

From Composite of Continuous Mappings is Continuous and Linear Function is Continuous, we therefore have:

$F_{a X + b}$ is continuous

in the case $a < 0$.

$\blacksquare$