author: Tyson Liddell
tags: mathematics probability
Let $$X$$ be a random variable on $$(\Om, \mathcal{F}, \prob)$$. Then for $$a,b \in \R$$, the function $$Y := aX + b$$ is a random variable. Indeed, if $$a > 0$$ then for any $$x \in \R$$ we have
$$ { Y \leq x } = { aX + b \leq x } = { \om \in \Om : aX(\om) + b \leq x } = \left{ \om \in \Om : X(\om) \leq \frac{x-b}{a} \right}. $$
When $$a < 0$$ we have
$$ \begin{align*} { Y \leq x } &= \left{ \om \in \Om : X(\om) \geq \frac{x-b}{a} \right} \ &= \left{ \om \in \Om : X(\om) \lt \frac{x-b}{a} \right}^C \ &= \left(\bigcup_{i\in\N} \left{ \om \in \Om : X(\om) \leq \frac{x-b}{a} - \frac{1}{n} \right}\right)^C. \end{align*} $$
Finally, when $$a = 0$$
$$ { Y \leq x } = \left{ \om \in \Om : b \leq x \right} \in { \Om, \emptyset }. $$
In all of the cases above, $${ Y \leq x } \in \mathcal{F}$$, making $$Y$$ a random variable.
If $$a > 0$$, then
$$ \begin{align*} F_Y(y) &= \prob(Y \leq y) \ &= \prob\left(\left{ \om \in \Om : X(\om) \leq \frac{y-b}{a} \right}\right) \ &= F_X\left(\frac{y-b}{a}\right), \end{align*} $$
and when $$a < 0$$ we have
$$ \begin{align} F_Y(y) &= \prob(Y \leq y) \ &= \prob\left( \left(\bigcup_{i\in\N} \left{ \om \in \Om : X(\om) \leq \frac{y-b}{a} - \frac{1}{n} \right}\right)^C\right) \ &= 1 - \prob\left(\bigcup_{i\in\N} \left{ \om \in \Om : X(\om) \leq \frac{y-b}{a} - \frac{1}{n} \right}\right) \ &= 1 - \lim_{n\to\infty} \prob\left( \left{ \om \in \Om : X(\om) \leq \frac{y-b}{a} - \frac{1}{n} \right}\right) \tag{1}\label{continuity} \ &= 1 - \lim_{n\to\infty} F_X\left(\frac{y-b}{a} - \frac{1}{n}\right)\ &= 1 - \lim_{x \uparrow \frac{y-b}{a}} F_X(x). \tag{2}\label{limit_exists} \end{align} $$
At $$(\ref{continuity})$$ we used the continuity property of $$\prob$$ and at $$(\ref{limit_exists})$$ the fact that $$F_X$$ is monotonic implies that its left and right sided limits exist at every point. Note that in the special case where $$F_X$$ is continuous, the above simplifies to $$F_Y(y) = 1 - F_X\left(\frac{y-b}{a}\right)$$.
Since $$Y$$ is a CDF, it must be right continuous. This is a direct consequence of the continuity property of $$\prob$$:
$$ \lim_{h \to 0^{+}} F_Y(y + h) = \lim_{h \to 0^{+}} \prob({Y \leq y + h}) = \lim_{n \to \infty} \prob({Y \leq y + 1/n}) = \prob({Y \leq y}) = F_Y(y), $$
since $${Y \leq y + 1/n}$$ is a decreasing sequence of events whose intersection is equal to $${ Y \leq y}$$. Even with this result, I didn't find it obvious that the expression at $$(\ref{limit_exists})$$ should be right continuous. However, recalling that $$a < 0$$, the continuity property of $$\prob$$ implies that,
$$ \begin{align*} \lim_{h \to 0^{+}} F_Y(y + h) &= \lim_{h \to 0^{+}} \left[ 1 - \lim_{g \to 0^{-}} F_X\left(\frac{y + h - b}{a} + g\right) \right] \ &= 1 - \lim_{h \to 0^{+}} \lim_{g \to 0^{-}} \prob\left( \left{X \leq \frac{y + h - b}{a} + g\right} \right) \ &= 1 - \lim_{h \to 0^{+}} \prob\left(\left{X \lt \frac{y + h - b}{a}\right}\right) \ &= 1 - \lim_{h \to 0^{-}} \prob\left(\left{X \lt \frac{y - b}{a} + \frac{h}{|a|}\right}\right) \ &= 1 - \lim_{h \to 0^{-}} \prob\left(\left{X \leq \frac{y - b}{a} + h\right}\right) \ &= 1 - \lim_{h \to 0^{-}}F_X\left(\frac{y - b}{a} + h\right) \ &= F_Y(y) \end{align*} $$
which confirms, explicitly, that $$(\ref{limit_exists})$$ is right continuous.
The choice of inequality in the definition $$F_X(x) = \prob(X \leq x)$$ determines the left/right continuity of cumulative distribution functions. Indeed, if we define $$\widehat{F}_X(x) = \prob(X \lt x)$$, the continuity property of $$\prob$$ guarantees that the left and right limits satisfy
$$ \lim_{y \to x^{-}}F_X(y) = \prob({X < x}) = \lim_{y \to x^{-}}\widehat{F}X(y) \quad\text{and}\quad \lim{y \to x^{+}}F_X(y) = \prob({X \leq x}) = \lim_{y \to x^{+}}\widehat{F}_X(y) $$
Therefore, the fact that the CDF is left or right continuous is determined by whether or not we choose to define it as $$F_X$$ or $$\widehat{F}_X$$. Since $$F_X$$ was chosen, it is right continuous.