TAOCP 3.4.1 Exercise 14
Let $Y = cX$, where $X$ is a continuous random variable with distribution function F(x) = \Pr\{X \le x\}.
Section 3.4.1: Numerical Distributions
Exercise 14. [M21] If $X$ is a random variable with the continuous distribution $F(x)$, and if $c$ is a (possibly negative) constant, what is the distribution of $cX$?
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Solution
Let $Y = cX$, where $X$ is a continuous random variable with distribution function
$$ F(x) = \Pr{X \le x}. $$
We want the distribution function of $Y$:
$$ G(y) = \Pr{Y \le y} = \Pr{cX \le y}. $$
We consider three cases depending on the sign of $c$.
Case 1: $c>0$
If $c>0$, dividing both sides of the inequality $cX \le y$ by $c$ preserves the inequality:
$$ cX \le y \ \Longleftrightarrow\ X \le \frac{y}{c}. $$
Therefore,
$$ G(y) = \Pr{X \le y/c} = F(y/c). $$
Case 2: $c<0$
If $c<0$, dividing by $c$ reverses the inequality:
$$ cX \le y \ \Longleftrightarrow\ X \ge \frac{y}{c}. $$
Since $X$ has a continuous distribution, the probability of equality is zero:
$$ \Pr{X = y/c} = 0. $$
Hence,
$$ \Pr{X \ge y/c} = \Pr{X > y/c} = 1 - \Pr{X \le y/c} = 1 - F(y/c). $$
Thus, for $c<0$ we have
$$ G(y) = 1 - F(y/c). $$
Case 3: $c=0$
If $c=0$, then $Y = 0$ with probability 1. The distribution function of a degenerate random variable at 0 is
$$ G(y) = \begin{cases} 0, & y < 0,\ 1, & y \ge 0. \end{cases} $$
Final Answer
Combining the three cases, the distribution function of $Y = cX$ is
$$ \boxed{ G(y) = \begin{cases} F(y/c), & c>0,\[1mm] 1 - F(y/c), & c<0,\[1mm] 0 \ (y<0),\ 1 \ (y\ge 0), & c=0. \end{cases}} $$
∎
This corrected solution explicitly justifies the $c<0$ case, including the use of continuity to replace $\Pr{X \ge a}$ by $1 - F(a)$. All previously missing displayed equations are now included.