Transformations of Random Variables

Study guide: Exam MAS-I

A Transformation of a Random Variable produces a new variable $Y = g(X)$ from an existing one. The reliable way to find its distribution is the CDF method: write $F_Y(y) = P(g(X) \leq y)$ , rearrange the inequality into an event about $X$ , evaluate with $F_X$ , then differentiate to get the density.

$F_Y(y) = P\bigl(g(X) \leq y\bigr)$

$f_Y(y) = \frac{d}{dy}F_Y(y)$

$f_Y(y) = f_X\bigl(g^{-1}(y)\bigr)\left|\frac{d}{dy}g^{-1}(y)\right|$

The third line is the change-of-variable shortcut, valid only when $g$ is strictly monotone on the support of $X$ . The Jacobian factor $|dg^{-1}/dy|$ is what most solutions forget.
If $g$ is not monotone (e.g. $Y = X^2$ with $X$ taking both signs), the shortcut fails — go back to the CDF method and collect every branch of $X$ that maps into the region.
Always carry the support through the transformation. Deriving a correct formula on the wrong interval is the most common way to lose the mark.
A linear transformation $Y = aX + b$ gives $E[Y] = aE[X] + b$ and $\text{Var}(Y) = a^2\text{Var}(X)$ directly, with no integration needed.
Insurance applications are transformations: $Y = (X-d)_+$ under a Deductible, $Y = \min(X, u)$ under a Benefit Limit, and $X' = (1+r)X$ under Inflation all reshape the Loss Random Variable into the Payment Random Variable.
If $E[g(X)]$ is all that is wanted, do not find $f_Y$ — use $E[g(X)] = \int g(x) f_X(x)\,dx$ directly.

Scaling an Exponential Loss for Inflation Losses follow $X \sim \text{Exponential}$ with mean 1000. Next year losses inflate by 10%, so $Y = 1.1X$ . Find the density of $Y$ .
Answer $g(x) = 1.1x$ is strictly increasing, with $g^{-1}(y) = y/1.1$ and $\frac{d}{dy}g^{-1}(y) = 1/1.1$ . Apply the change-of-variable formula:
 $\begin{align*} f_Y(y) &= f_X\!\left(\frac{y}{1.1}\right) \cdot \frac{1}{1.1} \\ &= \frac{1}{1000}e^{-y/1100} \cdot \frac{1}{1.1} \\ &= \frac{1}{1100}e^{-y/1100}, \quad y > 0 \end{align*}$ 
That is exponential with mean 1100 — scaling an exponential just scales its mean. Dropping the $1/1.1$ Jacobian would have left a density that does not integrate to 1.

The Probability Integral Transform $X$ is continuous with strictly increasing CDF $F_X$ . Show that $U = F_X(X)$ is $\text{Uniform}(0,1)$ .
Answer Use the CDF method. For $0 < u < 1$ :
 $\begin{align*} F_U(u) &= P\bigl(F_X(X) \leq u\bigr) \\ &= P\bigl(X \leq F_X^{-1}(u)\bigr) \\ &= F_X\bigl(F_X^{-1}(u)\bigr) \\ &= u \end{align*}$ 
$F_U(u) = u$ on $(0,1)$ is exactly the uniform CDF. Run backwards, for any CDF $F$ that is strictly increasing where $0 < F < 1$ , $X = F^{-1}(U)$ has CDF $F$ — so uniform random numbers can be turned into samples from such a distribution, the basis of simulation.

A Non-Monotone Transformation $X \sim \text{Uniform}(-1, 1)$ , so $f_X(x) = 1/2$ on $(-1,1)$ . Find the density of $Y = X^2$ .
Answer $g(x) = x^2$ is not monotone on $(-1,1)$ , so the shortcut does not apply. Use the CDF method, keeping both branches of $X$ that produce $Y \leq y$ :
 $\begin{align*} F_Y(y) &= P(X^2 \leq y) \\ &= P(-\sqrt{y} \leq X \leq \sqrt{y}) \\ &= \frac{2\sqrt{y}}{2} \\ &= \sqrt{y}, \quad 0 < y < 1 \end{align*}$ 
Differentiating: $f_Y(y) = \frac{1}{2\sqrt{y}}, \quad 0 < y < 1$ Note the support collapsed from $(-1,1)$ to $(0,1)$ — squaring folds the negative half onto the positive half.